When the light of wavelength `400 nm` is incident on a metal surface of work function `2.3 eV`, photoelectrons are emitted. A fasted photoelectron combines with a `He^(2+)` ion to form `He^(+)` ion in its third excited state and a photon is emitted in this process. Then the energy of the photon emitted during combination is

Options

A. `4.2 eV`
B. `5.7 eV`
C. `6.5 eV`
D. `6.8 eV`

Explanation

To solve the problem step by step, we need to calculate the energy of the incident light, the kinetic energy of the emitted photoelectron, and the energy of the photon emitted during the combination of the photoelectron with the He²⁺ ion.

### Step 1: Calculate the energy of the incident light
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 4.14 \times 10^{-15} \) eV·s
- \( c \) (speed of light) = \( 3 \times 10^8 \) m/s
- \( \lambda \) (wavelength) = \( 400 \) nm = \( 400 \times 10^{-9} \) m

Calculating \( E \):
\[ E = \frac{(4.14 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{400 \times 10^{-9} \text{ m}} \]

\[ E = \frac{(1.242 \times 10^{-6} \text{ eV·m})}{400 \times 10^{-9} \text{ m}} \]

\[ E = 3.105 \text{ eV} \approx 3.1 \text{ eV} \]

### Step 2: Calculate the kinetic energy of the emitted photoelectron
The kinetic energy (KE) of the emitted photoelectron can be calculated using the equation:
\[ KE = E - \phi \]
Where:
- \( E \) = energy of the incident light = \( 3.1 \) eV
- \( \phi \) (work function) = \( 2.3 \) eV

Calculating \( KE \):
\[ KE = 3.1 \text{ eV} - 2.3 \text{ eV} \]
\[ KE = 0.8 \text{ eV} \]

### Step 3: Calculate the energy of the He⁺ ion in the third excited state
The energy of the He⁺ ion in the nth excited state can be calculated using the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2} \]
Where:
- \( Z \) (atomic number of Helium) = 2
- \( n \) (principal quantum number for the third excited state) = 4

Calculating \( E_n \):
\[ E_n = -\frac{2^2 \cdot 13.6 \text{ eV}}{4^2} \]
\[ E_n = -\frac{4 \cdot 13.6 \text{ eV}}{16} \]
\[ E_n = -\frac{54.4 \text{ eV}}{16} \]
\[ E_n = -3.4 \text{ eV} \]

### Step 4: Calculate the energy of the emitted photon during the combination
The energy of the emitted photon during the combination of the photoelectron with the He²⁺ ion is equal to the kinetic energy of the photoelectron plus the energy of the He⁺ ion in the third excited state:
\[ E_{\text{photon}} = KE + |E_n| \]
Where:
- \( KE = 0.8 \) eV
- \( |E_n| = 3.4 \) eV

Calculating \( E_{\text{photon}} \):
\[ E_{\text{photon}} = 0.8 \text{ eV} + 3.4 \text{ eV} \]
\[ E_{\text{photon}} = 4.2 \text{ eV} \]

### Final Answer
The energy of the photon emitted during the combination is **4.2 eV**.

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