Q.The work function of caesium metal is $2.14\ \text{eV}$. When light of frequency $6 \times 10^{14}\ \text{Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the

When a hydrogen atom absorbs energy from a colliding electron, the electron in the atom jumps from its ground state ($n=1$) to a higher energy level. The energy levels of hydrogen are given by $E_n = -\frac{13.6}{n^2}$ eV. The energy difference between the ground state ($n=1$) and any excited state ($n$) is $\Delta E = 13.6 \left(1 - \frac{1}{n^2}\right)$ eV. The bombarding electron provides exactly 12.75 eV of energy, so we need to find which $n$ this corresponds to.

The number of spectral lines emitted when the atom de-excites depends on how many distinct downward jumps are possible. If the atom is excited to level $n$, the number of spectral lines is the number of ways to go from $n$ down to $1$, which is $\frac{n(n-1)}{2}$.

Let’s work through the steps.

1. Find the excited state reached. The ground state energy is $E_1 = -13.6$ eV. The energy of the $n$th level is $E_n = -\frac{13.6}{n^2}$ eV. The energy absorbed to go from $n=1$ to $n$ is:

Set this equal to 12.75 eV:

Divide both sides by 13.6:

So:

Hence $n^2 = \frac{1}{0.0625} = 16$, so $n = 4$.

The atom is excited to the $n=4$ level.

2. Determine the possible transitions.

   From $n=4$, the electron can fall to any lower level: $n=3$, $n=2$, or $n=1$. From $n=3$, it can fall to $n=2$ or $n=1$. From $n=2$, it can fall to $n=1$. Each distinct jump emits a photon of a specific wavelength, producing a spectral line.

   The total number of distinct spectral lines is the number of pairs of levels $(n_i, n_f)$ with $n_i > n_f$, starting from $n=4$ down to $n=1$. This is simply the number of combinations of 4 levels taken 2 at a time:

Alternatively, list them: $4 \to 3$, $4 \to 2$, $4 \to 1$, $3 \to 2$, $3 \to 1$, $2 \to 1$ — that’s 6 lines.

3. Confirm the energy matches exactly. The energy difference $E_4 - E_1 = 13.6(1 - 1/16) = 13.6 \times 15/16 = 12.75$ eV, which matches the bombarding energy exactly. So the atom is excited precisely to $n=4$, not ionized (ionization requires 13.6 eV). Thus, only transitions within the atom occur.

Concept & Intuition

In the Bohr model of the hydrogen atom, an electron’s total energy is $E_n = -\frac{13.6}{n^2}$ eV. But kinetic energy is positive and exactly equal to the magnitude of the total energy (since $E = KE + PE$, and $PE = -2 \times KE$ for a Coulomb force). So $KE_n = \frac{13.6}{n^2}$ eV.

The key: kinetic energy scales as $1/n^2$. The “excited states” are counted starting from the ground state ($n=1$) as the first. So the third excited state means $n=4$, and the fourth excited state means $n=5$. The ratio is then simply $(1/4^2) : (1/5^2) = 25 : 16$.

Step-by-step reasoning

1. Identify the quantum numbers

   • Ground state: $n=1$ (first energy level)
   • First excited state: $n=2$
   • Second excited state: $n=3$
   • Third excited state: $n=4$
   • Fourth excited state: $n=5$
   Watch out

   A common mistake is to think “third excited” means $n=3$. Remember: “excited” counts above the ground state, so the $k$-th excited state has $n = k+1$.

2. Write the kinetic energy formula

   For hydrogen, the kinetic energy in the $n$-th orbit is

(derived from $KE = \frac{1}{2}mv^2$ and $v \propto 1/n$).

3. Compute the ratio

So the ratio is $25 : 16$.

4. Match with options Option (C) is $25 : 16$.

The key idea is that in the Bohr model, the radius of the $n$-th orbit is proportional to $n^2$. So the difference between successive radii grows as we go to higher orbits. The condition given — that this difference equals the radius of a lower orbit — sets up a simple quadratic in $n$.

Let’s work through it step by step.

1. Write the radii in terms of $n$. In the Bohr model, the radius of the $n$-th orbit is

where $a_0$ is the Bohr radius (a constant). So we have:

2. Translate the given condition into an equation. The problem says:

Substituting the expressions:

Since $a_0 \neq 0$, we can divide through by $a_0$:

3. Simplify the left-hand side.

So the equation becomes:

4. Expand and solve the quadratic.

Cancel the $+1$ on both sides:

Bring all terms to one side:

Factor:

So $n = 0$ or $n = 4$.

5. Interpret the result physically. The principal quantum number $n$ is a positive integer starting from 1. $n = 0$ is not allowed. Hence the only valid solution is

In the Bohr model the electron's energy is quantised:

The ground state is $n=1$; an "excited state" is any state above it. So the first excited state is $n=2$ and the second excited state is $n=3$ (a common error is to call $n=1$ the first excited state — that is the ground state).

1. Energies (magnitudes).

2. Ratio.

So the ratio of the energies in the first and second excited states is $9:4$.

The concept here is the Bohr model of the hydrogen atom, where electrons occupy discrete energy levels given by $E_n = -\frac{13.6\,\text{eV}}{n^2}$, with $n=1,2,3,\dots$. When an electron jumps from a lower to a higher energy level, it must absorb a photon whose energy exactly matches the difference between those levels. The trick is correctly identifying which $n$ corresponds to the "second excited state"—a common pitfall is confusing "excited state" numbering with the principal quantum number.

Let’s work through it step by step:

1. Identify the initial and final states.

   The ground state is $n=1$. The "second excited state" means the state with the second-highest energy above the ground state. The first excited state is $n=2$, the second excited state is $n=3$. So the electron goes from $n=1$ to $n=3$.

2. Write the energy formula for hydrogen.

This gives negative energies because the electron is bound; the more negative, the lower the energy.

3. Calculate the energy of the ground state.

4. Calculate the energy of the second excited state.

5. Find the energy difference (photon energy). The photon must supply exactly the gap:

Rounded to two decimal places, this is $12.09\,\text{eV}$.

6. Match with the options. The options are: (A) 3.4 eV, (B) 10.2 eV, (C) 12.09 eV, (D) 12.75 eV. Our result is exactly 12.09 eV, which corresponds to option (C).

The energy levels of an electron in a hydrogen atom are quantized, meaning the electron can only exist at specific discrete energy values. These energy levels are described by the principal quantum number, $n$, where $n$ can be any positive integer ($1, 2, 3, \dots$).

The lowest possible energy state for an electron is called the ground state, which corresponds to $n=1$. Any state with $n > 1$ is called an excited state. The excited states are numbered sequentially:

• The first excited state corresponds to $n=2$.
• The second excited state corresponds to $n=3$.
• The third excited state corresponds to $n=4$.
• And so on.

In general, the $k$-th excited state corresponds to the principal quantum number $n = k+1$.

Let's apply this understanding to the problem.

1. Identify the principal quantum number ($n$):

   The question asks for the energy of an electron in the fourth excited state. Following our definition, the $k$-th excited state corresponds to $n = k+1$.

   For the fourth excited state, $k=4$, so the principal quantum number is $n = 4+1 = 5$.

   Watch out

   A common mistake is to assume that the "fourth excited state" means $n=4$. Remember that $n=1$ is the ground state, so the first excited state is $n=2$, the second is $n=3$, and the third is $n=4$. Therefore, the fourth excited state must be $n=5$.

2. Calculate the energy ($E_n$):

   Now, substitute $n=5$ into the energy formula for the hydrogen atom:

3. Compare with the given options:

   The calculated energy for the fourth excited state is $-0.544~\text{eV}$. Let's check the given options:

   (A) $-0.85~\text{eV}$ (This corresponds to $n=4$, which is the third excited state: $E_4 = -13.6/4^2 = -13.6/16 = -0.85~\text{eV}$)

   (B) $-1.70~\text{eV}$

   (C) $0$

   (D) $-0.425~\text{eV}$

   Our calculated value of $-0.544~\text{eV}$ is not among the provided options. This indicates a potential discrepancy in the question or options. However, based on the correct physical definition of excited states, the energy for the fourth excited state is indeed $-0.544~\text{eV}$.

The key concept here is the photoelectric effect: a photon can eject an electron from a metal only if its energy is at least as large as the metal’s work function (the minimum energy needed to remove an electron). The Hα photon’s energy is fixed by the hydrogen energy levels, so the maximum work function a metal can have and still emit photoelectrons is exactly that photon energy.

We find the photon energy from the hydrogen atom’s energy levels:

1. Recall the hydrogen energy formula The energy of a level with principal quantum number $n$ is

This comes from the Bohr model; the negative sign means the electron is bound.

2. Calculate the energies for n=3 and n=2

3. Find the photon energy from the transition The photon energy equals the difference between the initial and final levels:

This is the Hα photon energy.

4. Apply the photoelectric condition For photoelectrons to be emitted, the photon energy must be at least the work function $\phi$:

The maximum work function that still allows emission is when equality holds:

The relevant concept here is the inverse proportionality between photon energy and wavelength: $E_{\text{photon}} = \frac{hc}{\lambda}$. When an atom drops from a higher energy level to a lower one, the emitted photon’s energy equals the difference between the two levels. So if we know the energy difference for one transition, we can scale it to find the wavelength for another.

Let’s work through it step by step.

1. First transition: from $2E$ to $E$. The energy difference is

This photon has wavelength $\lambda$, so

2. Second transition: from $\frac{5}{3}E$ to $E$. The energy difference is

Let the new wavelength be $\lambda'$. Then

3. Relate the two: From step 1, $E = hc/\lambda$. Substitute this into step 2:

Cancel $hc$ (nonzero) from both sides:

Invert both sides:

First transition: $2E \to E$ releases a photon of energy

Second transition: $\tfrac{5}{3}E \to E$ releases

Since $\lambda \propto \dfrac{1}{\Delta E}$:

The orbital angular momentum of an electron in a hydrogen atom is quantized. For a given principal quantum number $n$, the orbital angular momentum is $\sqrt{l(l+1)} \frac{h}{2\pi}$, where $l$ can be $0, 1, \dots, n-1$. But the problem gives a simpler clue: the increase is exactly $\frac{h}{2\pi}$. That is the fundamental unit of angular momentum in quantum mechanics — the reduced Planck constant $\hbar = \frac{h}{2\pi}$.

When the angular momentum increases by exactly one unit of $\hbar$, the azimuthal quantum number $l$ must increase by 1. In the ground state ($n=1$), the only possible $l$ is $0$. So after absorbing energy, the electron jumps to a state where $l=1$. The smallest $n$ that allows $l=1$ is $n=2$. Therefore, the transition is from $n=1$ to $n=2$.

Now, the energy of a hydrogen atom in the $n$th state is $E_n = -\frac{13.6}{n^2}\ \text{eV}$. For the ground state ($n=1$), $E_1 = -13.6\ \text{eV}$. For $n=2$, $E_2 = -\frac{13.6}{4} = -3.4\ \text{eV}$.