Electrons from n = 2 to n= 1 in hydrogen atom is made to fall on a metal surface with work function 1.2ev. The maximum velocity of photo electrons emitted is nearly equal to

Options

A. `6 xx 10^(5)` m/s
B. `3xx 10^(5)` m/s
C. `2 xx 10^(5)` m/s
D. `18 xx 10^(5)` m/s

Answer

To solve the problem, we need to calculate the maximum velocity of photoelectrons emitted when electrons transition from the n=2 to n=1 energy level in a hydrogen atom and then fall on a metal surface with a work function of 1.2 eV.

### Step-by-Step Solution:

1. **Calculate the Energy Difference (E)**:
The energy of an electron in a hydrogen atom at a given principal quantum number \( n \) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
For \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
The energy difference \( E \) when an electron transitions from \( n=2 \) to \( n=1 \) is:
\[ E = E_1 - E_2 = (-13.6) - (-3.4) = -13.6 + 3.4 = -10.2 \, \text{eV} \]

2. **Consider the Work Function (Φ)**:
The work function of the metal is given as \( \Phi = 1.2 \, \text{eV} \).

3. **Calculate the Kinetic Energy (KE)**:
The kinetic energy of the emitted photoelectron can be calculated using the energy difference and the work function:
\[ KE = E - \Phi = 10.2 \, \text{eV} - 1.2 \, \text{eV} = 9.0 \, \text{eV} \]

4. **Convert Kinetic Energy to Joules**:
To find the maximum velocity, we need to convert the kinetic energy from electron volts to joules. We know:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore:
\[ KE = 9.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.44 \times 10^{-18} \, \text{J} \]

5. **Use the Kinetic Energy Formula**:
The kinetic energy of an electron is also given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron (\( m \approx 9.1 \times 10^{-31} \, \text{kg} \)). Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]

6. **Substituting Values**:
\[ v = \sqrt{\frac{2 \times 1.44 \times 10^{-18} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]

7. **Calculating the Velocity**:
\[ v = \sqrt{\frac{2.88 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{3.16 \times 10^{12}} \approx 1.78 \times 10^6 \, \text{m/s} \]

8. **Final Approximation**:
Rounding this value gives approximately:
\[ v \approx 1.8 \times 10^6 \, \text{m/s} \approx 18 \times 10^5 \, \text{m/s} \]

### Conclusion:
The maximum velocity of the photoelectrons emitted is nearly equal to \( 18 \times 10^5 \, \text{m/s} \).

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