Radiation corresponding to the transition n=4 to n=2 in hydrogen atoms falls on a certain metal (work function=2.5 eV). The maximum kinetic energy of the photo-electrons will be:

Options

A. (a) `0.55` eV
B. (b) `2.55` eV
C. (c) `4.45` eV
D. (d) None of these

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the wavelength (λ) of the radiation corresponding to the transition from n=4 to n=2 in hydrogen.
We will use the Rydberg formula:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant (\( R = 1.1 \times 10^5 \, \text{cm}^{-1} \))
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 = 2 \) (final energy level)
- \( n_2 = 4 \) (initial energy level)

Substituting the values:

\[ \frac{1}{\lambda} = 1.1 \times 10^5 \times 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the fractions:

\[ \frac{1}{\lambda} = 1.1 \times 10^5 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.1 \times 10^5 \left( \frac{4 - 1}{16} \right) = 1.1 \times 10^5 \times \frac{3}{16} \]

Now calculate \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = 1.1 \times 10^5 \times \frac{3}{16} = \frac{3.3 \times 10^5}{16} \approx 20625 \, \text{cm}^{-1} \]

Now, take the reciprocal to find \( \lambda \):

\[ \lambda \approx \frac{1}{20625} \approx 4.84 \times 10^{-5} \, \text{cm} \]

Convert to nanometers:

\[ \lambda \approx 484 \, \text{nm} \]

### Step 2: Calculate the energy (E) of the radiation using the wavelength.
The energy can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) (speed of light) = \( 3 \times 10^{10} \, \text{cm/s} \)

Using \( hc = 1240 \, \text{eV nm} \):

\[ E = \frac{1240}{484} \approx 2.56 \, \text{eV} \]

### Step 3: Calculate the maximum kinetic energy (K.E.) of the photoelectrons.
The maximum kinetic energy of the photoelectrons can be calculated using:

\[ K.E. = E - \text{Work Function} \]

Given the work function of the metal is \( 2.5 \, \text{eV} \):

\[ K.E. = 2.56 \, \text{eV} - 2.5 \, \text{eV} = 0.06 \, \text{eV} \]

### Final Answer:
The maximum kinetic energy of the photoelectrons is approximately **0.06 eV**.

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