The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 400 mm is 500 mV. When the incident wavelength is changed to a new value, the stopping potential is found to be 800 mV. New wavelength is about-

Options

A. 365m
B. 250 nm
C. 640 nm
D. 340 nm

Explanation

To solve the problem step by step, we will use the photoelectric effect equation and the concept of stopping potential.

### Step 1: Understand the relationship between stopping potential and wavelength
The stopping potential \( V_s \) is related to the maximum kinetic energy of the emitted photoelectrons. The equation is given by:

\[ E V_s = h \nu - \phi \]

Where:
- \( E \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) C),
- \( V_s \) is the stopping potential,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s),
- \( \nu \) is the frequency of the light,
- \( \phi \) is the work function of the material.

### Step 2: Convert wavelength to frequency
The frequency \( \nu \) can be expressed in terms of wavelength \( \lambda \):

\[ \nu = \frac{c}{\lambda} \]

Where \( c \) is the speed of light (\( 3 \times 10^8 \) m/s).

### Step 3: Write the equations for both stopping potentials
For the first case, with \( \lambda_1 = 400 \) nm and \( V_{s1} = 500 \) mV:

\[ E \cdot 500 \times 10^{-3} = h \left( \frac{c}{400 \times 10^{-9}} \right) - \phi \]

For the second case, with \( V_{s2} = 800 \) mV and \( \lambda_2 \) being the new wavelength:

\[ E \cdot 800 \times 10^{-3} = h \left( \frac{c}{\lambda_2} \right) - \phi \]

### Step 4: Set up the equations
We can express the two equations as:

1. \( 500 \times 10^{-3} E = \frac{hc}{400 \times 10^{-9}} - \phi \) (Equation 1)
2. \( 800 \times 10^{-3} E = \frac{hc}{\lambda_2} - \phi \) (Equation 2)

### Step 5: Eliminate the work function \( \phi \)
To find \( \lambda_2 \), we can eliminate \( \phi \) by subtracting Equation 1 from Equation 2:

\[ (800 \times 10^{-3} E - 500 \times 10^{-3} E) = \left( \frac{hc}{\lambda_2} - \frac{hc}{400 \times 10^{-9}} \right) \]

This simplifies to:

\[ 300 \times 10^{-3} E = hc \left( \frac{1}{\lambda_2} - \frac{1}{400 \times 10^{-9}} \right) \]

### Step 6: Solve for \( \lambda_2 \)
Rearranging gives:

\[ \frac{1}{\lambda_2} = \frac{300 \times 10^{-3} E}{hc} + \frac{1}{400 \times 10^{-9}} \]

Now substituting values for \( E \), \( h \), and \( c \):

- \( E = 1.6 \times 10^{-19} \) C
- \( h = 6.626 \times 10^{-34} \) J·s
- \( c = 3 \times 10^8 \) m/s

Calculating \( \frac{hc}{400 \times 10^{-9}} \):

\[ \frac{hc}{400 \times 10^{-9}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}} \approx 4.9695 \times 10^{-19} \text{ J} \]

Now substituting back to find \( \lambda_2 \):

Using the above values, we can calculate \( \lambda_2 \) and find it to be approximately \( 364.77 \) nm.

### Final Answer
The new wavelength \( \lambda_2 \) is approximately **365 nm**.

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