Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. T

Q: Why do we use $n=5$ for the fourth excited state?

The ground state is $n=1$. The first excited state is $n=2$, the second is $n=3$, and so on. Therefore, the fourth excited state corresponds to $n=5$.

Q: How do we calculate the energy of the photon emitted during recombination?

Energy must be conserved. The initial energy is the kinetic energy of the free electron ($K_{\text{max}}$) plus the potential energy (which is zero at infinity). The final energy is the bound state energy ($E_5$) plus the energy of the emitted photon. So, $E_{\text{photon}} = K_{\text{max}} - E_5$.

Q: Why do we only consider the maximum kinetic energy electron for recombination?

The problem specifically states 'A maximum energy electron combines with an $\alpha$-particle'. This means we must use $K_{\text{max}}$ calculated from the photoelectric equation.

Q: Why do we use $Z=2$ for the energy levels?

The $\alpha$-particle is a helium nucleus. When it captures one electron, it becomes a hydrogen-like $\text{He}^+$ ion, which has an atomic number $Z=2$.

Q: Do we need to check transitions from $n=3$ to lower states?

The energy of the $n=3$ state is $-6.04 \text{ eV}$ and $n=2$ is $-13.6 \text{ eV}$. The transition $3 \rightarrow 2$ emits a photon of $7.56 \text{ eV}$, which is already greater than 4 eV. Any transition to $n=1$ will be even larger. So, we don't need to check them.

Question

Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing $\alpha$-particles. A maximum energy electron combines with an $\alpha$-particle to form a $	ext{He}^+$ ion, emitting a single photon in this process. $	ext{He}^+$ ions thus formed are in their fourth excited state. Find the energies in eV of the photons lying in the 2 to 4 eV range, that are likely to be emitted during and after the combination.
[Take $h = 4.14 	imes 10^{-15}$ eV-s]

Sigma Percentile · Accepted Answer

### The Photoelectric Kickoff

We start with a photon of wavelength $400 	ext{ nm}$ striking a metal surface. Using the energy formula $E = \frac{hc}{\lambda}$, we find the incident energy to be $3.1 	ext{ eV}$.

The metal has a work function of $1.9 	ext{ eV}$. By Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron is simply the difference: $K_{	ext{max}} = 3.1 	ext{ eV} - 1.9 	ext{ eV} = 1.2 	ext{ eV}$.

### The Recombination Event

This energetic electron now enters a region filled with $\alpha$-particles (helium nuclei, $Z=2$). It gets captured, forming a $	ext{He}^+$ ion.

The problem states it lands in the fourth excited state. Remember, the ground state is $n=1$, so the fourth excited state is $n=5$.

The energy of this state is given by Bohr's formula: $E_5 = -13.6 \frac{Z^2}{n^2} = -13.6 \frac{2^2}{5^2} \approx -2.2 	ext{ eV}$.

When the free electron (with $1.2 	ext{ eV}$ of kinetic energy) falls into this $-2.2 	ext{ eV}$ bound state, it must shed the excess energy as a photon.

The energy of this recombination photon is $\Delta E = 1.2 	ext{ eV} - (-2.2 	ext{ eV}) = 3.4 	ext{ eV}$. This perfectly falls within our target range of $2 	ext{ eV}$ to $4 	ext{ eV}$!

### The De-excitation Cascade

Now, the $	ext{He}^+$ ion is sitting in the $n=5$ state, but it won't stay there. It will cascade down to lower energy levels, emitting more photons.

Let's calculate the energies of the lower states:
$$E_4 = -13.6 \frac{4}{16} = -3.4 	ext{ eV}$$
$$E_3 = -13.6 \frac{4}{9} = -6.04 	ext{ eV}$$
$$E_2 = -13.6 \frac{4}{4} = -13.6 	ext{ eV}$$

Now we check the possible transitions from $n=5$:
- $5 ightarrow 4$: $\Delta E = -2.2 - (-3.4) = 1.2 	ext{ eV}$ (Too low)
- $5 ightarrow 3$: $\Delta E = -2.2 - (-6.04) = 3.84 	ext{ eV}$ (**In range!**)
- $5 ightarrow 2$: $\Delta E = -2.2 - (-13.6) = 11.4 	ext{ eV}$ (Too high)

Next, we check transitions from $n=4$:
- $4 ightarrow 3$: $\Delta E = -3.4 - (-6.04) = 2.64 	ext{ eV}$ (**In range!**)
- $4 ightarrow 2$: $\Delta E = -3.4 - (-13.6) = 10.2 	ext{ eV}$ (Too high)

Any transitions to $n=1$ or from $n=3$ downwards will yield energies much greater than $4 	ext{ eV}$.

### The Final Verdict

Gathering all our in-range photons, we have one from the initial recombination and two from the subsequent cascade.

The energies of the photons lying in the $2 	ext{ eV}$ to $4 	ext{ eV}$ range are **$3.4 	ext{ eV}$**, **$3.84 	ext{ eV}$**, and **$2.64 	ext{ eV}$**.

Visualized Solution

$E = \frac{h c}{λ}$

$K_{max} = E - W$

$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$

$Δ E_{combination} = K_{max} - E_{5}$

$E_{n} for He^{+}$

$Δ E = E_{i} - E_{f}$

$Δ E = E_{i} - E_{f}$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Photoelectric Kickoff

The Recombination Event

The De-excitation Cascade

The Final Verdict

Similar Questions

An excited $He^{+}$ ion emits two photons in succession, with wavelengths $108.5 nm$ and $30.4 nm$ , in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $λ$ , energy $E = \frac{1240 eV}{λ (in nm)}$ ]

Comprehension Passage

The quantum number $n$ of the state finally populated in He $^{+}$ ions is

The wavelength of light emitted in the visible region by He $^{+}$ ions after collisions with H atoms is

The ratio of the kinetic energy of the $n = 2$ electron for the H atom to that of He $^{+}$ ion is

Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms fall on $He^{+}$ ions in $n = 1$ and $n = 2$ states. The possible transition of helium ions as they absorb energy from the radiation is

A free electron of $2.6 eV$ energy collides with a $H^{+}$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. ( $h = 6.6 \times 1 0^{- 34} Js$ )

A $He^{+}$ ion is in its first excited state. Its ionisation energy is

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=λhc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kmax​=E−W

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​=−13.6n2Z2​ eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔEcombination​=Kmax​−E5​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​ for He+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=Ei​−Ef​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=Ei​−Ef​

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Photoelectric Kickoff

The Recombination Event

The De-excitation Cascade

The Final Verdict

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The quantum number n of the state finally populated in He+ ions is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The wavelength of light emitted in the visible region by He+ ions after collisions with H atoms is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ratio of the kinetic energy of the n=2 electron for the H atom to that of He+ ion is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A He+ ion is in its first excited state. Its ionisation energy is

$E = \frac{h c}{λ}$

$K_{max} = E - W$

$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$

$Δ E_{combination} = K_{max} - E_{5}$

$E_{n} for He^{+}$

$Δ E = E_{i} - E_{f}$

$Δ E = E_{i} - E_{f}$

The quantum number $n$ of the state finally populated in He $^{+}$ ions is

The wavelength of light emitted in the visible region by He $^{+}$ ions after collisions with H atoms is

The ratio of the kinetic energy of the $n = 2$ electron for the H atom to that of He $^{+}$ ion is

A $He^{+}$ ion is in its first excited state. Its ionisation energy is