Q.Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the time period of an electron in the ground state of hydrogen atom is $1.5 \times 10^{-16}$ s, then the time period of the electron in the third excited state of hydrogen atom is (A) $2.4 \times 10^{-15}$ s (B) $9.6 \times 10^{-15}$ s (C) $6.4 \times 10^{-15}$ s (D) $4.8 \times 10^{-15}$ s

   ›Reveal solutionSolution

   The time period of an electron in a hydrogen atom scales as $n^3$, so the third excited state ($n=4$) has a period $4^3 = 64$ times the ground-state period, giving $9.6 \times 10^{-15}$ s.

   The key idea is that the electron in a hydrogen atom moves in a circular orbit (in the Bohr model) with a period proportional to the cube of the principal quantum number $n$. This comes from combining the Bohr quantization condition with the centripetal force law. Once we know the scaling, we can directly compute the period for any excited state from the ground-state value.

   Why $T \propto n^3$?

   In the Bohr model, the radius of the $n$-th orbit is $r_n = n^2 a_0$, and the orbital speed is $v_n = v_1 / n$, where $v_1$ is the speed in the ground state. The period is $T_n = 2\pi r_n / v_n$. Substituting:

   $T_n = 2\pi (n^2 a_0) / (v_1 / n) = (2\pi a_0 / v_1) \, n^3 = T_1 \, n^3$.

   So the period scales as $n^3$.

   Now, step by step:

   1. Identify the state number.

      The ground state corresponds to $n=1$. The "third excited state" means the electron has been excited three times above the ground state: $n = 1 + 3 = 4$. (Be careful: the first excited state is $n=2$, second is $n=3$, third is $n=4$.)

   2. Apply the scaling law.

      Since $T_n = T_1 \cdot n^3$, for $n=4$:

   $T_4 = T_1 \cdot 4^3 = T_1 \cdot 64.$

   3. Plug in the given value. $T_1 = 1.5 \times 10^{-16}$ s, so …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.When an electron at rest is accelerated through an electric potential, the de Broglie wavelength associated with the electron is $\lambda$. For de Broglie wavelength associated with the electron to become $\dfrac{2\lambda}{3}$, the percentage increase in the potential to be applied is (A) 75 (B) 225 (C) 125 (D) 150

   ›Reveal solutionSolution

   The de Broglie wavelength of an electron is inversely proportional to the square root of the accelerating potential. To reduce the wavelength to $\frac{2}{3}$ of its original value, the potential must increase by a factor of $\frac{9}{4}$, which corresponds to a 125% increase. The correct option is (C).

   The key idea here is the relationship between de Broglie wavelength and accelerating potential for a charged particle. When an electron is accelerated from rest through a potential difference $V$, it gains kinetic energy equal to $eV$. This kinetic energy is non-relativistic for typical potentials in such problems, so we can write:

   $\frac{1}{2} m v^2 = eV$

   The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant and $m$ is the electron mass. Combining these, we get a direct proportionality between $\lambda$ and $1/\sqrt{V}$.

   Let’s work through it step by step.

   1. Write the expression for de Broglie wavelength in terms of potential. From $eV = \frac{1}{2}mv^2$, we get $v = \sqrt{\frac{2eV}{m}}$. Substituting into $\lambda = \frac{h}{mv}$:

   $\lambda = \frac{h}{m \sqrt{\frac{2eV}{m}}} = \frac{h}{\sqrt{2meV}}$

   So $\lambda \propto \frac{1}{\sqrt{V}}$. This is the central proportionality — double the potential, and the wavelength shrinks by $\sqrt{2}$.

   2. Set up the ratio for the new wavelength. Let the original potential be $V_1$, giving $\lambda_1 = \lambda$. The new potential $V_2$ gives $\lambda_2 = \frac{2\lambda}{3}$. Using the proportionality:

   $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$

   Substitute the given ratio:

   $\frac{2}{3} = \sqrt{\frac{V_1}{V_2}}$

   3. Solve for the ratio of potentials. Square both sides:

   $\frac{4}{9} = \frac{V_1}{V_2} \quad \Rightarrow \quad V_2 = \frac{9}{4} V_1$

   So the new potential is $2.25$ times the original. …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.If the speed of the electron in an orbit of hydrogen atom is $\dfrac{1}{548}$ times the speed of light in vacuum, then the angular momentum of the electron in this orbit is (h – Planck’s constant) (A) $\dfrac{2h}{\pi}$ (B) $\dfrac{4h}{\pi}$ (C) $\dfrac{h}{2\pi}$ (D) $\dfrac{3h}{\pi}$

   ›Reveal solutionSolution

   The speed of an electron in a hydrogen atom's orbit is related to the principal quantum number $n$ and the fine-structure constant $\alpha$. By using the given speed to find $n$, we can then determine the angular momentum using Bohr's quantization rule. The angular momentum is $\boxed{\frac{2h}{\pi}}$.

   The problem connects the speed of an electron in a hydrogen atom's orbit to its angular momentum. This requires understanding the Bohr model, specifically the formulas for electron speed and angular momentum in a given orbit.

   The key idea is that both the speed of the electron and its angular momentum are quantized, meaning they can only take on discrete values determined by the principal quantum number $n$. We are given the electron's speed, which allows us to identify the specific orbit (i.e., find the value of $n$). Once $n$ is known, we can directly calculate the angular momentum using Bohr's quantization condition.

   1. Relate electron speed to the principal quantum number: In the Bohr model for a hydrogen atom, the speed of an electron in the $n$-th orbit, $v_n$, is given by the formula:

   $v_n = \frac{1}{n} \frac{e^2}{2\epsilon_0 h}$

   This formula can be expressed more compactly using the fine-structure constant, $\alpha$. The fine-structure constant is defined as $\alpha = \frac{e^2}{2\epsilon_0 hc}$, where $c$ is the speed of light in vacuum.
   Substituting this into the speed formula, we get:
   

   $v_n = \frac{1}{n} \alpha c$

   The approximate value of the fine-structure constant is $\alpha \approx \frac{1}{137}$.
   
   > [!FORMULA]
   > The speed of an electron in the $n$-th orbit of a hydrogen atom is $v_n = \frac{\alpha c}{n}$.
   

   2. Determine the principal quantum number ($n$):

   We are given that the speed of the electron is $\frac{1}{548}$ times the speed of light in vacuum. So, $v = \frac{1}{548} c$.

   Equating this with the formula for $v_n$: …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the difference in the frequencies of the first and second lines of Lyman series of hydrogen atom is $f$, then the difference in frequencies of the first and second lines of Balmer series of hydrogen atom is (A) $\dfrac{3f}{4}$ (B) $f$ (C) $\dfrac{7f}{20}$ (D) $\dfrac{9f}{16}$

   ›Reveal solutionSolution

   The key idea is that the frequency of any spectral line in hydrogen is proportional to the difference of two inverse-square terms, and the Lyman and Balmer series share the same lower-level pattern. The required ratio simplifies to 7/20, so the answer is (C).

   The problem asks for the difference in frequencies between the first and second lines of the Balmer series, given that the same difference for the Lyman series is $f$. Since both series come from the hydrogen atom, we can use the Rydberg formula for frequency:

   $\nu = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

   where $R$ is the Rydberg constant (in frequency units), $n_1$ is the lower energy level, and $n_2 > n_1$ is the upper level. The trick is to write the differences for each series and compare them algebraically — the constant $R$ cancels out.

   1. Identify the lines in the Lyman series Lyman series: $n_1 = 1$.
      • First line: $n_2 = 2$ → $\nu_{L1} = R\left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$.
      • Second line: $n_2 = 3$ → $\nu_{L2} = R\left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R\left(1 - \frac{1}{9}\right) = \frac{8R}{9}$. Their difference:

   $f = \nu_{L2} - \nu_{L1} = \frac{8R}{9} - \frac{3R}{4} = R\left( \frac{32}{36} - \frac{27}{36} \right) = \frac{5R}{36}.$

   2. Identify the lines in the Balmer series Balmer series: $n_1 = 2$.
      • First line: $n_2 = 3$ → $\nu_{B1} = R\left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R\left( \frac{1}{4} - \frac{1}{9} \right) = R\left( \frac{9-4}{36} \right) = \frac{5R}{36}$.
      • Second line: $n_2 = 4$ → $\nu_{B2} = R\left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R\left( \frac{1}{4} - \frac{1}{16} \right) = R\left( \frac{4-1}{16} \right) = \frac{3R}{16}$. Their difference:

   $\Delta \nu_B = \nu_{B2} - \nu_{B1} = \frac{3R}{16} - \frac{5R}{36}.$

   3. Compute the Balmer difference in terms of $f$ Find a common denominator for $\frac{3}{16}$ and $\frac{5}{36}$: LCM of 16 and 36 is 144.

   $\frac{3}{16} = \frac{27}{144}, \quad \frac{5}{36} = \frac{20}{144}.$

   So

   $\Delta \nu_B = R\left( \frac{27}{144} - \frac{20}{144} \right) = \frac{7R}{144}.$

   From step 1, $f = \frac{5R}{36} = \frac{20R}{144}$. Therefore,

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.When a photosensitive material is illuminated by photons of energy $3.1~\text{eV}$, the stopping potential of the photoelectrons is $1.7~\text{V}$. When the same photosensitive material is illuminated by photons of energy $2.5~\text{eV}$, the stopping potential of the photoelectrons is (A) $1.8~\text{V}$ (B) $1.4~\text{V}$ (C) $1.1~\text{V}$ (D) $1.3~\text{V}$

   ›Reveal solutionSolution

   The photoelectric equation relates photon energy, work function, and stopping potential. Using the two given conditions, the work function cancels out, giving the second stopping potential as $1.1~\text{V}$.

   The core idea here is the photoelectric effect equation:

   $E_{\text{photon}} = \phi + eV_s$

   where $\phi$ is the work function (the minimum energy needed to eject an electron) and $V_s$ is the stopping potential (the voltage that just stops the most energetic photoelectrons). The stopping potential multiplied by the electron charge $e$ gives the maximum kinetic energy of the photoelectrons in eV.

   Since the same material is used in both cases, the work function $\phi$ is constant. That means we can write two equations and subtract them to eliminate $\phi$, directly finding the unknown stopping potential.

   1. Write the equation for the first case Photon energy $E_1 = 3.1~\text{eV}$, stopping potential $V_{s1} = 1.7~\text{V}$:

   $3.1 = \phi + 1.7$

   (Here $eV_s$ is numerically equal to $V_s$ in eV because $e=1$ in these units.)

   2. Write the equation for the second case Photon energy $E_2 = 2.5~\text{eV}$, unknown stopping potential $V_{s2}$:

   $2.5 = \phi + V_{s2}$

   3. Subtract the two equations to eliminate $\phi$:

   $(3.1 - 2.5) = (1.7 - V_{s2})$

   $0.6 = 1.7 - V_{s2}$ …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The ratio of frequencies of second line of Lyman series and third line of Balmer series of hydrogen atom is (A) $360 : 174$ (B) $27 : 5$ (C) $5 : 36$ (D) $800 : 189$

   ›Reveal solutionSolution

   The frequency of emitted light during an electron transition in a hydrogen atom is determined by the Rydberg formula. By identifying the initial and final energy levels for the second line of the Lyman series ($n_i=3 \to n_f=1$) and the third line of the Balmer series ($n_i=5 \to n_f=2$), we find their frequency ratio to be $\boxed{800 : 189}$.

   Concept and Intuition

   When an electron in a hydrogen atom transitions from a higher energy level to a lower one, it emits a photon. The energy of this photon corresponds precisely to the difference in energy between the two levels. This emitted energy manifests as light of a specific frequency (and wavelength), giving rise to the characteristic spectral lines of hydrogen.

   The energy levels of a hydrogen atom are quantized, meaning electrons can only exist in discrete orbits, each with a specific energy. The energy of an electron in the $n$-th orbit is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$, where $n$ is the principal quantum number ($n=1, 2, 3, \dots$).

   When an electron jumps from an initial energy level $n_i$ to a final energy level $n_f$ (where $n_i > n_f$), the energy of the emitted photon is $\Delta E = E_{n_i} - E_{n_f}$. According to Planck's relation, this energy is also equal to $h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the emitted photon.

   Combining these ideas, we can derive a general formula for the frequency of the emitted photon:

   $\nu = \frac{\Delta E}{h} = \frac{E_{n_i} - E_{n_f}}{h}$

   Substituting the expression for $E_n$:

   $\nu = \frac{1}{h} \left( -\frac{R_H h c}{n_i^2} - \left(-\frac{R_H h c}{n_f^2}\right) \right)$

   $\nu = R_H c \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

   Here, $R_H$ is the Rydberg constant and $c$ is the speed of light. This is the Rydberg formula for frequency, which we will use to solve the problem.

   Different series of spectral lines correspond to different final energy levels ($n_f$):

   • Lyman series: Transitions to $n_f = 1$ (ultraviolet region).
   • Balmer series: Transitions to $n_f = 2$ (visible region).
   • Paschen series: Transitions to $n_f = 3$ (infrared region). And so on. Within each series, the "first line" corresponds to the smallest possible jump (e.g., $n_i=2 \to n_f=1$ for Lyman), the "second line" to the next smallest jump (e.g., $n_i=3 \to n_f=1$ for Lyman), and so forth.

   Step-by-Step Solution

   1. Identify the transition for the second line of the Lyman series. The Lyman series corresponds to transitions where the electron falls to the ground state, $n_f = 1$.
      • The first line of the Lyman series is for $n_i = 2 \to n_f = 1$.
      • The second line of the Lyman series is for $n_i = 3 \to n_f = 1$. Using the frequency formula:

   $\nu_L = R_H c \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = R_H c \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$

   $\nu_L = R_H c \left( 1 - \frac{1}{9} \right) = R_H c \left( \frac{9-1}{9} \right) = R_H c \left( \frac{8}{9} \right)$

   2. Identify the transition for the third line of the Balmer series. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.The energy gap between conduction and valence bands of silicon is (A) $5.4~\text{eV}$ (B) $1.1~\text{eV}$ (C) $0.7~\text{eV}$ (D) $1.4~\text{eV}$

   ›Reveal solutionSolution

   Silicon is a semiconductor with a characteristic band gap that determines its electrical properties at room temperature. The energy gap is $1.1~\text{eV}$.

   The band gap of a material is the minimum energy required to excite an electron from the valence band (where electrons are bound to atoms) to the conduction band (where they are free to move and conduct electricity). This fundamental property distinguishes insulators, semiconductors, and conductors.

   For insulators, the band gap is very large (typically $> 3~\text{eV}$), making thermal excitation at room temperature negligible. Conductors have overlapping bands with no gap. Semiconductors occupy the middle ground: their band gaps are small enough that a significant number of electrons can be thermally excited at room temperature, yet large enough that the material isn't a good conductor in its pure state.

   Silicon is the archetypal semiconductor, and its band gap is one of the most important numbers in solid-state physics and electronics. Let me walk through why each option does or doesn't make sense:

   1. Option (A): $5.4~\text{eV}$ – This is far too large for a semiconductor. A gap this wide would make silicon behave like an insulator at room temperature (thermal energy $k_B T \approx 0.026~\text{eV}$ at $300~\text{K}$ is negligible compared to $5.4~\text{eV}$). Diamond, an insulator, has a band gap around $5.5~\text{eV}$, which is close to this value.

   2. Option (B): $1.1~\text{eV}$ – This is the correct value for silicon at room temperature. It's small enough that thermal excitation produces a modest number of charge carriers, giving silicon its semiconducting properties. This value is precisely why silicon dominates the electronics industry. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The ratio of the centripetal accelerations of the electron in two successive orbits of hydrogen is 81:16. Due to a transition between these two states, the angular momentum of the electron changes by (h – Planck’s constant) (A) $\dfrac{h}{3\pi}$ (B) $\dfrac{3h}{\pi}$ (C) $\dfrac{h}{2\pi}$ (D) $\dfrac{2h}{\pi}$

   ›Reveal solutionSolution

   The centripetal acceleration in Bohr's model scales as $a \propto 1/n^4$, so a ratio of 81:16 (i.e. $3^4:2^4$) corresponds to the two successive orbits $n=2$ and $n=3$. The angular momentum change for a transition between them is $\Delta L = \hbar = h/(2\pi)$, so the correct option is (C).

   Concept & Intuition

   In Bohr's hydrogen atom, the electron moves in circular orbits with quantized angular momentum $L = n\hbar$, where $\hbar = h/(2\pi)$. The centripetal acceleration is $a = v^2/r$. Using Bohr's relations $v \propto 1/n$ and $r \propto n^2$, the acceleration scales as $a \propto 1/n^4$. So the given ratio of accelerations lets us pin down the two quantum numbers, and the angular momentum change follows directly from their difference.

   Step-by-step reasoning

   1. Centripetal acceleration in Bohr's model. For a hydrogen atom, the Coulomb force provides the centripetal force:

   $\frac{mv^2}{r} = \frac{ke^2}{r^2}$

   With $r_n \propto n^2$ and $v_n \propto 1/n$,

   $a_n = \frac{v_n^2}{r_n} \propto \frac{1/n^2}{n^2} = \frac{1}{n^4}.$

   2. Use the given ratio.

   $\frac{a_{n_1}}{a_{n_2}} = \frac{n_2^4}{n_1^4} = \frac{81}{16} = \frac{3^4}{2^4}$

   So $n_2/n_1 = 3/2$, and since these are stated to be successive orbits, the smallest integers satisfying this ratio are $n_1 = 2$ and $n_2 = 3$ (check: $2^4=16$, $3^4=81$, giving exactly $81:16$).

   3. Angular momentum change. …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.In hydrogen atom, the frequency of the photon emitted when an electron jumps from second orbit to first orbit is ‘f’. The frequency of the photon emitted when an electron jumps from third excited state to first excited state is (A) $\dfrac{f}{2}$ (B) $\dfrac{f}{4}$ (C) $\dfrac{f}{8}$ (D) $f$

   ›Reveal solutionSolution

   The key idea is that the frequency of emitted light in hydrogen is proportional to the difference of inverse squares of the principal quantum numbers. For the given transitions, the ratio of frequencies is 1/4, so the answer is f/4.

   The relevant concept is the Rydberg formula for hydrogen: the wavenumber (or frequency) of emitted light is proportional to $\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $n_i$ is the initial orbit and $n_f$ is the final orbit. The frequency $f$ is directly proportional to this difference because $E = hf$ and the energy difference between orbits follows the same pattern. So we don’t need to plug in constants — we just compare the two transitions.

   1. Identify the quantum numbers for the first transition. The electron jumps from the second orbit ($n_i = 2$) to the first orbit ($n_f = 1$). The frequency $f$ is proportional to:

   $f \propto \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 1 - \frac{1}{4} = \frac{3}{4}.$

   2. Identify the quantum numbers for the second transition. “Third excited state” means $n = 4$ (since ground state is $n=1$, first excited is $n=2$, second excited is $n=3$, third excited is $n=4$). “First excited state” means $n = 2$. So the transition is from $n_i = 4$ to $n_f = 2$. The frequency $f'$ is proportional to: $f' \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16}.$ …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
    Q.When a body is projected vertically up, at a point ‘P’ in its path, the ratio of potential energy to kinetic energy is $2:3$. If the same body is projected up with double the previous velocity, the ratio of potential energy to kinetic energy of the body, at the same point ‘P’ is (A) $2:25$ (B) $1:9$ (C) $2:9$ (D) $4:9$

    ›Reveal solutionSolution

    The ratio of potential to kinetic energy at a fixed point depends on the square of the projection speed. Using energy conservation, the new ratio becomes $2:9$, so option (C) is correct.

    The key idea here is that potential energy at a given point depends only on height, while kinetic energy depends on speed. When you change the projection speed, the total mechanical energy changes, but the potential energy at point P stays the same (since P is at a fixed height). So the ratio changes because the kinetic energy at P changes.

    Let’s work through it step by step.

    1. Set up the first case. Let the mass of the body be $m$, and let the initial projection speed be $u$. At point P, let the height be $h$ (so potential energy $U = mgh$). The ratio given is:

    $\frac{U}{K} = \frac{2}{3}$

    where $K$ is the kinetic energy at P. This means $U = \frac{2}{3}K$, or equivalently $K = \frac{3}{2}U$.

    2. Use energy conservation for the first case. Total mechanical energy at projection = total mechanical energy at P:

    $\frac{1}{2} m u^2 = U + K = U + \frac{3}{2}U = \frac{5}{2}U$

    So:

    $U = \frac{1}{5} m u^2$

    This tells us the potential energy at P in terms of the initial kinetic energy.

    3. Now consider the second case. The projection speed is doubled: $u' = 2u$. The new total energy is:

    $\frac{1}{2} m (2u)^2 = \frac{1}{2} m \cdot 4u^2 = 2 m u^2$

    At the same point P, the potential energy $U$ is unchanged (same height $h$), so $U = \frac{1}{5} m u^2$ as before.

    4. Find the new kinetic energy at P. By energy conservation at P:

    $\text{Total energy} = U + K'$

    $2 m u^2 = \frac{1}{5} m u^2 + K'$

    $K' = 2 m u^2 - \frac{1}{5} m u^2 = \frac{10}{5} m u^2 - \frac{1}{5} m u^2 = \frac{9}{5} m u^2$

    5. Compute the new ratio. $\frac{U}{K'} = \frac{\frac{1}{5} m u^2}{\frac{9}{5} m u^2} = \frac{1}{9}$ …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
    Q.Photoelectrons are emitted with maximum kinetic energies 1.2 eV and 3.6 eV when lights of wavelengths $\lambda$ and $\dfrac{\lambda}{2}$ respectively incident on a photosensitive material. The work function of the photosensitive material is (A) 1.2 eV (B) 2.4 eV (C) 3.6 eV (D) 4.8 eV

    ›Reveal solutionSolution

    Using Einstein’s photoelectric equation for two wavelengths, the work function is found by eliminating the unknown constant. The work function is 1.2 eV.

    The photoelectric effect tells us that when light of a certain frequency (or wavelength) strikes a metal surface, electrons are ejected if the photon energy exceeds the material’s work function. The excess energy appears as the kinetic energy of the emitted electron. Einstein’s photoelectric equation gives the maximum kinetic energy as:

    $K_{\text{max}} = h\nu - \phi$

    where $h\nu$ is the photon energy and $\phi$ is the work function. Since frequency $\nu = c/\lambda$, we can write it in terms of wavelength:

    $K_{\text{max}} = \frac{hc}{\lambda} - \phi$

    Here we have two different wavelengths, $\lambda$ and $\lambda/2$, producing two different maximum kinetic energies. The work function $\phi$ is a property of the material and remains the same in both cases. So we can set up two equations and solve for $\phi$.

    1. Write the equations for both cases. For wavelength $\lambda$, $K_1 = 1.2$ eV:

    $1.2 = \frac{hc}{\lambda} - \phi \quad \text{(1)}$

    For wavelength $\lambda/2$, $K_2 = 3.6$ eV:

    $3.6 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi \quad \text{(2)}$

    2. Eliminate the unknown $hc/\lambda$. Let $x = \frac{hc}{\lambda}$. Then equation (1) becomes:

    $1.2 = x - \phi \quad \Rightarrow \quad x = 1.2 + \phi$

    Equation (2) becomes:

    $3.6 = 2x - \phi$

    Substitute $x$ from the first into the second:

    $3.6 = 2(1.2 + \phi) - \phi$

    3. Solve for $\phi$. Expand:

    $3.6 = 2.4 + 2\phi - \phi = 2.4 + \phi$

    Subtract 2.4 from both sides: …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The additional energy that should be given to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is (A) four times initial energy (B) thrice the initial energy (C) equal to the initial energy (D) twice the initial energy

    ›Reveal solutionSolution

    The de-Broglie wavelength is inversely proportional to momentum, and for an electron, kinetic energy is proportional to the square of momentum. Halving the wavelength doubles the momentum, which quadruples the kinetic energy. The additional energy needed is therefore three times the initial energy.

    The key concept here is the de-Broglie relation and how kinetic energy scales with momentum for a non-relativistic particle.

    For an electron (or any particle), the de-Broglie wavelength is given by

    $\lambda = \frac{h}{p}$

    where $h$ is Planck’s constant and $p$ is the momentum.

    If the wavelength is halved, the momentum must double.

    Now, the kinetic energy $E$ of a non-relativistic electron is

    $E = \frac{p^2}{2m}$

    so energy is proportional to the square of momentum.

    Thus, doubling momentum quadruples the energy.

    The question asks for the additional energy, not the final energy — so we subtract the initial energy from the final energy.

    Let’s work through it step by step.

    1. Initial wavelength and momentum Let the initial wavelength be $\lambda_1 = 1\ \text{nm}$. The initial momentum is

    $p_1 = \frac{h}{\lambda_1}$

    The initial kinetic energy is

    $E_1 = \frac{p_1^2}{2m} = \frac{h^2}{2m \lambda_1^2}$

    2. Final wavelength and momentum The final wavelength is $\lambda_2 = 0.5\ \text{nm} = \frac{\lambda_1}{2}$. The final momentum is

    $p_2 = \frac{h}{\lambda_2} = \frac{h}{\lambda_1/2} = \frac{2h}{\lambda_1} = 2p_1$

    So momentum doubles.

    3. Final kinetic energy

    $E_2 = \frac{p_2^2}{2m} = \frac{(2p_1)^2}{2m} = \frac{4p_1^2}{2m} = 4E_1$

    So the final energy is four times the initial energy. …

    🔒 Start your 14-day free trial to see the full solution →