Q.Light of frequency $7.21 \times 10^{14}\ \text{Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5}\ \text{m/s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Showing the 12 most recent of 24 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.When photons of wavelength $4000~\text{\AA}$ are incident on a photosensitive material of cut-off wavelength $4800~\text{\AA}$, the stopping potential is $V$. If the same photons incident on another photosensitive material of cut off wavelength $6000~\text{\AA}$, then the stopping potential is (A) $1.5V$ (B) $0.5V$ (C) $4V$ (D) $2V$

   ›Reveal solutionSolution

   With fixed incident wavelength, $eV=hc\!\left(\tfrac1\lambda-\tfrac1{\lambda_0}\right)$; the new stopping potential works out to $2V$.

   Photoelectric equation with incident wavelength $\lambda=4000\,\text{\AA}$ and cut-off $\lambda_0$:

   $eV=\frac{hc}{\lambda}-\frac{hc}{\lambda_0}=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$

   First material, $\lambda_0=4800\,\text{\AA}$:

   $eV=hc\left(\frac{1}{4000}-\frac{1}{4800}\right)=hc\cdot\frac{1}{24000}$ …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The process of electron emission that takes place in a spark plug is (A) Field emission (B) Photoelectric emission (C) Thermionic emission (D) Secondary emission

   ›Reveal solutionSolution

   The intense electric field generated across the spark plug gap pulls electrons directly from the electrode surface, a process known as field emission, which initiates the spark. The correct option is (A).

   Concept and Intuition

   A spark plug's primary function is to ignite the air-fuel mixture in an internal combustion engine by creating an electrical spark. This spark is essentially a controlled electrical breakdown of the gas between two electrodes. For this breakdown to occur, electrons must first be emitted from the negative electrode (cathode) into the gas gap. These initial electrons are then accelerated by the strong electric field, collide with gas molecules, and ionize them, leading to an avalanche of charge carriers that forms the visible spark.

   The key to understanding the emission process lies in the conditions present in a spark plug:

   1. High Voltage: The ignition system applies tens of thousands of volts across a small gap (typically 0.6 to 1.5 mm).
   2. Intense Electric Field: This high voltage across a small gap creates an extremely strong electric field.
   3. Ambient Temperature: While the engine cylinder can get hot, the initial spark generation, especially during a cold start, does not rely on the electrodes being at a very high temperature.
   4. No External Light/Particles: There is no external source of high-energy light or bombarding particles to initiate the process.

   We need to identify which electron emission mechanism is best suited to these conditions.

   Step-by-step Analysis

   Let's examine each type of electron emission:

   1. Field Emission:

      • Concept: Field emission occurs when a very strong electric field (typically $10^7 \text{ V/m}$ to $10^9 \text{ V/m}$) is applied to a metal surface. This intense field distorts the potential energy barrier at the metal surface, making it thin enough for electrons to "tunnel" through it and escape the metal, even at low temperatures.
      • Relevance to Spark Plug: A spark plug generates extremely high voltages. For example, if $20,000 \text{ V}$ is applied across a $1 \text{ mm}$ gap, the average electric field is $E = \frac{V}{d} = \frac{20,000 \text{ V}}{1 \times 10^{-3} \text{ m}} = 2 \times 10^7 \text{ V/m}$. Local field enhancements at sharp points or irregularities on the electrode surface can make the actual field much higher, easily reaching the threshold for field emission. These emitted electrons are crucial for initiating the gas breakdown and forming the spark.
      • Conclusion: This is a highly plausible mechanism for initiating the spark.

   2. Photoelectric Emission:

      • Concept: Photoelectric emission is the ejection of electrons from a metal surface when light of sufficient frequency (and thus photon energy) strikes it.
      • Relevance to Spark Plug: There is no external light source of sufficient energy (like ultraviolet light) directed at the spark plug electrodes to cause electron emission.
      • Conclusion: This is not the mechanism.

   3. Thermionic Emission: …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.When monochromatic photons incident on a photosensitive material of work function $1.8~\text{eV}$, photoelectrons are emitted with a maximum kinetic energy of $2.2~\text{eV}$. If the frequency of the incident photons is doubled, then the maximum kinetic energy of the emitted photoelectrons is (A) $7.2~\text{eV}$ (B) $6.2~\text{eV}$ (C) $4.4~\text{eV}$ (D) $3.6~\text{eV}$

   ›Reveal solutionSolution

   Using Einstein’s photoelectric equation, the incident photon energy is found from the given work function and maximum kinetic energy; doubling the frequency doubles the photon energy, and the new kinetic energy is that doubled energy minus the work function, giving 6.2 eV.

   The key idea is Einstein’s photoelectric equation:

   $K_{\text{max}} = hf - \phi$

   where $K_{\text{max}}$ is the maximum kinetic energy of emitted electrons, $hf$ is the energy of an incident photon, and $\phi$ is the work function (the minimum energy needed to eject an electron).

   We are given $\phi = 1.8~\text{eV}$ and $K_{\text{max}} = 2.2~\text{eV}$ for the original frequency $f$. This lets us find the original photon energy. Then, if the frequency is doubled, the new photon energy is $h(2f) = 2hf$. The new kinetic energy is simply that new photon energy minus the same work function.

   ---

   1. Find the original photon energy From $K_{\text{max}} = hf - \phi$:

   $hf = K_{\text{max}} + \phi = 2.2~\text{eV} + 1.8~\text{eV} = 4.0~\text{eV}$

   2. Double the frequency The new photon energy is:

   $h(2f) = 2 \times (hf) = 2 \times 4.0~\text{eV} = 8.0~\text{eV}$

   3. Compute the new maximum kinetic energy Using the same equation with the new photon energy: $$$$

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The slope of the graph drawn by taking the frequency (in $10^{15}$ Hz) of light incident on a photosensitive material on x-axis and the stopping potential (in volt) on y-axis is nearly (A) 8.250 (B) 2.420 (C) 4.125 (D) 3.175

   ›Reveal solutionSolution

   The slope of the stopping potential vs. frequency graph in the photoelectric effect is $h/e$, and its numerical value is approximately $4.125 \times 10^{-15} \text{ V·s}$, which matches option (C).

   The photoelectric effect equation is the key here. Einstein’s photoelectric equation relates the stopping potential $V_0$, the frequency $f$ of incident light, the work function $\phi$, and Planck’s constant $h$:

   $eV_0 = hf - \phi$

   Here $e$ is the electron charge. Rearranging for $V_0$ as a function of $f$:

   $V_0 = \frac{h}{e} f - \frac{\phi}{e}$

   This is a straight line of the form $y = mx + c$, where $y = V_0$, $x = f$, the slope $m = h/e$, and the intercept $c = -\phi/e$. So the slope of the graph is simply Planck’s constant divided by the electronic charge.

   Now, the problem gives frequency in units of $10^{15}$ Hz and stopping potential in volts. That means the x-axis is actually $f \times 10^{-15}$ Hz — but the slope we compute must account for this scaling. Let’s work it out step by step.

   1. Recall the known constants Planck’s constant $h = 6.626 \times 10^{-34} \text{ J·s}$ Electron charge $e = 1.602 \times 10^{-19} \text{ C}$ The slope in SI units (V/Hz) is:

   $\frac{h}{e} = \frac{6.626 \times 10^{-34}}{1.602 \times 10^{-19}} \approx 4.135 \times 10^{-15} \text{ V/Hz}$

   2. Account for the x-axis scaling The x-axis uses frequency in units of $10^{15}$ Hz. That means if the actual frequency is $f$ Hz, the graph plots $f / 10^{15}$ on the x-axis. So the slope read from the graph (change in $V_0$ per unit change on the scaled axis) is:

   $\text{slope}_{\text{graph}} = \frac{\Delta V_0}{\Delta (f / 10^{15})} = 10^{15} \times \frac{\Delta V_0}{\Delta f} = 10^{15} \times \frac{h}{e}$

   3. Compute the numerical value Using $h/e \approx 4.135 \times 10^{-15}$ V/Hz: …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.When photons incident on a photosensitive material of work function $1.5$ eV, the maximum velocity of the emitted photoelectrons is $8 \times 10^5$ ms$^{-1}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9 \times 10^{-31}$ kg and charge of the electron $= 1.6 \times 10^{-19}$ C) (A) 1.8 V (B) 1.5 V (C) 2.1 V (D) 2.4 V

   ›Reveal solutionSolution

   The stopping potential is found by equating the maximum kinetic energy of the photoelectrons to the work done by the stopping voltage. Using $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$ and $K_{\text{max}} = e V_s$, we get $V_s \approx 1.8 \, \text{V}$, so the correct option is (A).

   The key idea here is that the stopping potential is the voltage that just barely stops the fastest photoelectrons from reaching the other electrode. That means the electrical work done on the electron (charge × voltage) exactly equals its maximum kinetic energy. So we don’t even need the work function or the photon energy — the maximum speed alone gives us the stopping potential directly.

   1. Write the relation between kinetic energy and stopping potential. The stopping potential $V_s$ is defined by:

   $e V_s = K_{\text{max}}$

   where $e = 1.6 \times 10^{-19} \, \text{C}$ is the electron charge and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons.

   2. Compute the maximum kinetic energy from the given speed. The maximum speed is $v_{\text{max}} = 8 \times 10^5 \, \text{m/s}$ and the electron mass is $m = 9 \times 10^{-31} \, \text{kg}$.

   $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} \times (9 \times 10^{-31}) \times (8 \times 10^5)^2$

   First square the speed: $(8 \times 10^5)^2 = 64 \times 10^{10} = 6.4 \times 10^{11}$.

   Then:

   $K_{\text{max}} = \frac{1}{2} \times 9 \times 10^{-31} \times 6.4 \times 10^{11} = \frac{1}{2} \times 57.6 \times 10^{-20} = 28.8 \times 10^{-20} \, \text{J}$

   So $K_{\text{max}} = 2.88 \times 10^{-19} \, \text{J}$.

   3. Convert this energy to electronvolts (optional but helpful for intuition). Since $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$, $K_{\text{max}} = \frac{2.88 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.8 \, \text{eV}$ …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.When photons incident on a photosensitive material of work function $1.5~\text{eV}$, the maximum velocity of the emitted photoelectrons is $8 \times 10^5~\text{ms}^{-1}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9 \times 10^{-31}~\text{kg}$ and charge of the electron $= 1.6 \times 10^{-19}~\text{C}$) (A) $2.1~\text{V}$ (B) $2.4~\text{V}$ (C) $1.5~\text{V}$ (D) $1.8~\text{V}$

   ›Reveal solutionSolution

   The stopping potential is found by equating the maximum kinetic energy of the photoelectrons to the work done by the stopping voltage. Using $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$ and $K_{\text{max}} = e V_s$, we get $V_s \approx 1.8~\text{V}$, so the correct option is (D).

   The key idea here is that the stopping potential is the voltage that just barely stops the fastest photoelectrons from reaching the other electrode. In the photoelectric effect, the maximum kinetic energy of the emitted electrons is given by Einstein’s equation:

   $K_{\text{max}} = hf - \phi$

   But we don’t know the photon frequency $f$ directly. However, we do know the maximum speed of the electrons, so we can compute $K_{\text{max}}$ from classical mechanics. Then, because the stopping potential $V_s$ is defined by $K_{\text{max}} = e V_s$ (the electron’s charge times the voltage), we can solve for $V_s$ directly — no need for the work function or photon energy at all in this step.

   ---

   1. Compute the maximum kinetic energy from the given speed. The maximum speed is $v_{\text{max}} = 8 \times 10^5~\text{m/s}$. The electron mass is $m = 9 \times 10^{-31}~\text{kg}$.

   $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} \times (9 \times 10^{-31}) \times (8 \times 10^5)^2$

   First square the speed: $(8 \times 10^5)^2 = 64 \times 10^{10} = 6.4 \times 10^{11}$.

   Then multiply:

   $\frac{1}{2} \times 9 \times 10^{-31} \times 6.4 \times 10^{11} = \frac{1}{2} \times 57.6 \times 10^{-20} = 28.8 \times 10^{-20}~\text{J}$

   So $K_{\text{max}} = 2.88 \times 10^{-19}~\text{J}$.

   2. Relate kinetic energy to stopping potential. The stopping potential $V_s$ is defined such that the work done by the electric field equals the maximum kinetic energy:

   $e V_s = K_{\text{max}}$

   where $e = 1.6 \times 10^{-19}~\text{C}$.

   Thus:

   $V_s = \frac{K_{\text{max}}}{e} = \frac{2.88 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.8~\text{V}$

   3. Check consistency with the work function (optional). …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.In a photoelectric experiment, the slope of the graph drawn between stopping potential along y-axis and frequency of incident radiation along x-axis is (Planck’s constant = $6.6 \times 10^{-34}$ Js) (A) $2.42 \times 10^{15}$ Js C$^{-1}$ (B) $10.56 \times 10^{-15}$ Js C$^{-1}$ (C) $4.125 \times 10^{-15}$ Js C$^{-1}$ (D) $6.25 \times 10^{-20}$ Js C$^{-1}$

   ›Reveal solutionSolution

   The slope of the stopping potential vs. frequency graph is Planck’s constant divided by the elementary charge, giving $4.125 \times 10^{-15}$ Js C$^{-1}$, so the correct option is (C).

   The key idea here is the photoelectric equation:

   $eV_0 = hf - \phi$

   where $V_0$ is the stopping potential, $f$ is the frequency, $h$ is Planck’s constant, $e$ is the electron charge, and $\phi$ is the work function.

   If you plot $V_0$ (y-axis) against $f$ (x-axis), the equation becomes

   $V_0 = \frac{h}{e} f - \frac{\phi}{e}$

   which is a straight line of the form $y = mx + c$.

   The slope $m$ is therefore $\frac{h}{e}$.

   We are given $h = 6.6 \times 10^{-34}$ Js, and the elementary charge $e = 1.6 \times 10^{-19}$ C (a standard constant you must recall).

   1. Identify the slope from the equation

      From $V_0 = \frac{h}{e} f - \frac{\phi}{e}$, the slope is $\frac{h}{e}$.

      No other quantity affects the slope — the work function only shifts the intercept.

   2. Plug in the numbers

   $\text{slope} = \frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19}} \ \text{Js C}^{-1}$

   3. Perform the division

   $\frac{6.6}{1.6} = 4.125$

   and for the powers of ten:

   $\frac{10^{-34}}{10^{-19}} = 10^{-15}$

   So the slope is $4.125 \times 10^{-15}$ Js C$^{-1}$. …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.In a photoelectric experiment, the slope of the graph drawn between stopping potential along y-axis and frequency of incident radiation along x-axis is (Planck’s constant = $6.6 \times 10^{-34}$ Js) (A) $6.25 \times 10^{-20}~\mathrm{JsC}^{-1}$ (B) $10.56 \times 10^{-15}~\mathrm{JsC}^{-1}$ (C) $4.125 \times 10^{-15}~\mathrm{JsC}^{-1}$ (D) $2.42 \times 10^{15}~\mathrm{JsC}^{-1}$

   ›Reveal solutionSolution

   The slope of the stopping potential vs. frequency graph is Planck’s constant divided by the elementary charge, $h/e$. Using $h = 6.6 \times 10^{-34}~\mathrm{Js}$ and $e = 1.6 \times 10^{-19}~\mathrm{C}$, the slope is $4.125 \times 10^{-15}~\mathrm{JsC}^{-1}$, which corresponds to option (C).

   The key concept here is the photoelectric equation:

   $eV_0 = hf - \phi$

   where $V_0$ is the stopping potential, $f$ is the frequency of incident light, $h$ is Planck’s constant, $e$ is the elementary charge, and $\phi$ is the work function. Rearranging:

   $V_0 = \frac{h}{e} f - \frac{\phi}{e}$

   This is a straight line of the form $y = mx + c$, with $y = V_0$, $x = f$, slope $m = h/e$, and intercept $c = -\phi/e$. So the slope of the graph is simply $h/e$, independent of the material.

   Now, let’s compute it step by step.

   1. Identify the known constants

      Planck’s constant: $h = 6.6 \times 10^{-34}~\mathrm{Js}$

      Elementary charge: $e = 1.6 \times 10^{-19}~\mathrm{C}$ (standard value, since not given explicitly but implied by the units in the options).

   2. Write the slope expression

      Slope $= \dfrac{h}{e} = \dfrac{6.6 \times 10^{-34}}{1.6 \times 10^{-19}}~\mathrm{Js\,C^{-1}}$

   3. Perform the division

      First, divide the numbers: $6.6 / 1.6 = 4.125$

      Then, divide the powers of ten: $10^{-34} / 10^{-19} = 10^{-15}$

      So slope $= 4.125 \times 10^{-15}~\mathrm{Js\,C^{-1}}$

   4. Match with the options …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If Planck’s constant is $6.63 \times 10^{-34}$ Js, then the slope of a graph drawn between cut off voltage and frequency of incident light in a photoelectric experiment is (A) $4.14 \times 10^{-15}$ Vs (B) $19.776 \times 10^{-15}$ Vs (C) $2.198 \times 10^{-15}$ Vs (D) $1.337 \times 10^{-15}$ Vs

   ›Reveal solutionSolution

   The slope of the cut‑off voltage vs. frequency graph equals Planck’s constant divided by the electron charge, $h/e$. Using $h = 6.63 \times 10^{-34}\ \text{Js}$ and $e = 1.6 \times 10^{-19}\ \text{C}$, the slope is $4.14 \times 10^{-15}\ \text{Vs}$, which matches option (A).

   ---

   The key idea comes from Einstein’s photoelectric equation:

   $eV_0 = hf - \phi$

   Here $V_0$ is the cut‑off (stopping) voltage, $f$ is the frequency of incident light, $h$ is Planck’s constant, $e$ is the electron charge, and $\phi$ is the work function of the metal. Rearranging:

   $V_0 = \frac{h}{e} f - \frac{\phi}{e}$

   This is a straight line of the form $y = mx + c$ when we plot $V_0$ (on the y‑axis) against $f$ (on the x‑axis). The slope $m$ is therefore $h/e$, a universal constant independent of the metal. So we just need to compute $h/e$ with the given $h$.

   ---

   1. Identify the slope

      From $V_0 = \frac{h}{e}f - \frac{\phi}{e}$, the slope is $\frac{h}{e}$.

   2. Write down the given constants

      • Planck’s constant: $h = 6.63 \times 10^{-34}\ \text{J·s}$
      • Electron charge: $e = 1.6 \times 10^{-19}\ \text{C}$ (standard value; not given but essential)

   3. Compute the slope

   $\frac{h}{e} = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} = \frac{6.63}{1.6} \times 10^{-15}$

   $\frac{6.63}{1.6} = 4.14375 \approx 4.14$ …

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10. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.The work functions of two photosensitive metal surfaces A and B are in the ratio 2:3. If x and y are the slopes of the graphs drawn between the stopping potential and frequency of incident light for the surfaces A and B respectively, then x : y = (A) 1 : 1 (B) 2 : 3 (C) 4 : 9 (D) 2 : 5

    ›Reveal solutionSolution

    The slope of the stopping potential vs. frequency graph is always $h/e$ (Planck’s constant divided by electron charge), independent of the work function. Therefore the ratio of slopes for any two surfaces is 1:1.

    Concept & Intuition

    The photoelectric effect equation is $eV_s = hf - \phi$, where $V_s$ is the stopping potential, $f$ the frequency, $\phi$ the work function, $h$ Planck’s constant, and $e$ the electron charge. Rearranging:

    $V_s = \frac{h}{e} f - \frac{\phi}{e}.$

    This is a straight line with slope $h/e$ and intercept $-\phi/e$. The slope depends only on fundamental constants $h$ and $e$, not on the metal’s work function. So no matter what the work functions are, the slopes of the $V_s$-vs-$f$ graphs are identical for all photosensitive surfaces.

    Step-by-step reasoning

    1. Write the photoelectric equation in linear form The stopping potential $V_s$ satisfies

    $eV_s = hf - \phi \quad \Rightarrow \quad V_s = \frac{h}{e}f - \frac{\phi}{e}.$

    This is of the form $y = mx + c$ with $x = f$, $y = V_s$.

    2. Identify the slope The slope $m$ is the coefficient of $f$:

    $m = \frac{h}{e}.$

    It contains only Planck’s constant $h$ and the elementary charge $e$ — both universal constants.

    3. Apply to surfaces A and B For surface A, slope $x = h/e$. …

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11. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.When light radiation of frequency $6.97 \times 10^{14}$ Hz is incident on a metal surface, then the electrons are ejected from the surface with a maximum speed of $6.6 \times 10^5$ ms$^{-1}$. The threshold frequency of the metal surface is (take $h = 6.6 \times 10^{-34}$ Js; mass of electron $= 9 \times 10^{-31}$ kg) (A) $5 \times 10^{14}$ Hz (B) $4 \times 10^{14}$ Hz (C) $3 \times 10^{14}$ Hz (D) $1 \times 10^{14}$ Hz

    ›Reveal solutionSolution

    The photoelectric equation links the incident photon energy to the electron’s kinetic energy and the work function. Using the given data, the threshold frequency comes out to be $4 \times 10^{14}$ Hz, which is option (B).

    The core idea here is Einstein’s photoelectric equation. When a photon strikes a metal surface, its energy is used first to overcome the binding energy (the work function) of the electron, and any leftover energy becomes the electron’s kinetic energy. The threshold frequency is the minimum frequency needed to just eject an electron — at that frequency, the kinetic energy is zero.

    We are given the incident frequency, the maximum speed of the ejected electrons, and the constants. So we can compute the kinetic energy, subtract it from the incident photon energy, and find the work function. From the work function, the threshold frequency follows directly.

    1. Write Einstein’s photoelectric equation The energy of an incident photon is $h\nu$. Part of it goes into the work function $\phi = h\nu_0$ (where $\nu_0$ is the threshold frequency), and the rest becomes the maximum kinetic energy of the ejected electron:

    $h\nu = h\nu_0 + \frac{1}{2} m v_{\text{max}}^2$

    2. Plug in the given numbers

       • $h = 6.6 \times 10^{-34}$ Js
       • $\nu = 6.97 \times 10^{14}$ Hz
       • $m = 9 \times 10^{-31}$ kg
       • $v_{\text{max}} = 6.6 \times 10^5$ m/s

       First, the incident photon energy:

    $h\nu = (6.6 \times 10^{-34})(6.97 \times 10^{14}) = 4.6002 \times 10^{-19} \text{ J}$

    Next, the maximum kinetic energy:

    $\frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} (9 \times 10^{-31}) (6.6 \times 10^5)^2$

    Compute $v_{\text{max}}^2 = (6.6 \times 10^5)^2 = 4.356 \times 10^{11}$ …

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12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.A photon released by the transition of an electron from the second excited state to the ground state of Hydrogen atom is incident on the surface of a metal of work function 3.1 eV. The de Broglie wavelength of the most energetic electron emitted from that metal surface is nearly (A) $2.6\,\text{\AA}$ (B) $4\,\text{\AA}$ (C) $6\,\text{\AA}$ (D) $7\,\text{\AA}$

    ›Reveal solutionSolution

    The key idea is to find the kinetic energy of the most energetic photoelectron by subtracting the work function from the photon energy (which comes from the hydrogen transition), then use the de Broglie relation to get its wavelength. The result is about 4 Å, so option (B) is correct.

    ---

    Concept & Intuition

    This problem combines two classic ideas:

    1. Atomic transitions in hydrogen — the photon energy equals the difference between two energy levels.
    2. Photoelectric effect — the maximum kinetic energy of an emitted electron is $K_{\text{max}} = hf - \phi$, where $\phi$ is the work function. Then, for that most energetic electron, we find its de Broglie wavelength: $\lambda = h / p$, and since the electron is non‑relativistic here, $p = \sqrt{2m_e K}$.

    The trick is to keep units consistent (eV, Å, etc.) and to remember that the “second excited state” means $n=3$ (ground is $n=1$).

    ---

    Step‑by‑step solution

    1. Identify the hydrogen transition The ground state is $n=1$. The second excited state is $n=3$ (first excited is $n=2$). The energy levels of hydrogen are

    $E_n = -\frac{13.6\ \text{eV}}{n^2}.$

    So

    $E_1 = -13.6\ \text{eV}, \quad E_3 = -\frac{13.6}{9}\ \text{eV} \approx -1.51\ \text{eV}.$

    2. Photon energy from the transition The photon energy is the difference:

    $E_{\text{photon}} = E_3 - E_1 = (-1.51) - (-13.6) = 12.09\ \text{eV}.$

    (More precisely, $13.6 \times (1 - 1/9) = 13.6 \times 8/9 \approx 12.09\ \text{eV}$.)

    3. Maximum kinetic energy of the photoelectron The work function of the metal is $\phi = 3.1\ \text{eV}$. By Einstein’s photoelectric equation:

    $K_{\text{max}} = E_{\text{photon}} - \phi = 12.09 - 3.1 = 8.99\ \text{eV} \approx 9.0\ \text{eV}.$

    4. Convert kinetic energy to joules $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$, so

    $K = 9.0 \times 1.602 \times 10^{-19} \approx 1.442 \times 10^{-18}\ \text{J}.$

    5. Find the electron’s momentum For a non‑relativistic electron (9 eV is tiny compared to its rest energy 511 keV),

    $p = \sqrt{2 m_e K},$

    with $m_e = 9.11 \times 10^{-31}\ \text{kg}$.

    $$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.442 \times 10^{-18}} \approx \sqrt{2.627 \times 10^{-48}} …$$

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