A metallic surface is irradiated with light energy E , of wave length `lamda` giving out electrons with maximum velocity 2m/s . The threshold wavelength of the metal surface is `3lamda` The same surface would release electrons with maximum velocity of 4m/s if irradiated by light of wavelength (nm) (m is mass of electrons)

Options

A. `(9m)/(hc)`
B. `(hc)/(9m)`
C. `(mh)/(9c)`
D. `(Elamda)/(9m)`

Answer

To solve the problem step by step, we will use the concepts of the photoelectric effect, including the work function, kinetic energy of emitted electrons, and the relationship between energy, wavelength, and frequency.

### Step 1: Understand the given data
- The initial wavelength of light is \( \lambda \).
- The maximum velocity of emitted electrons is \( V_{max1} = 2 \, \text{m/s} \).
- The threshold wavelength is \( \lambda_0 = 3\lambda \).
- The maximum velocity of emitted electrons when a different wavelength is used is \( V_{max2} = 4 \, \text{m/s} \).

### Step 2: Calculate the work function
The work function \( \phi \) can be calculated using the threshold wavelength:
\[ \phi = \frac{hc}{\lambda_0} = \frac{hc}{3\lambda} \]

### Step 3: Apply the photoelectric equation for the first case
Using the photoelectric equation:
\[ E = \phi + KE \]
Where:
- \( E = \frac{hc}{\lambda} \) (energy of the incident photon)
- \( KE = \frac{1}{2} mv_{max1}^2 = \frac{1}{2} m (2)^2 = 2m \)

Substituting these into the equation:
\[ \frac{hc}{\lambda} = \frac{hc}{3\lambda} + 2m \]

### Step 4: Rearranging the equation
Now, rearranging the equation gives:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = 2m \]
\[ \frac{hc}{\lambda} \left(1 - \frac{1}{3}\right) = 2m \]
\[ \frac{hc}{\lambda} \cdot \frac{2}{3} = 2m \]
\[ \frac{hc}{\lambda} = 3m \]

### Step 5: Apply the photoelectric equation for the second case
For the second case where the maximum velocity is \( 4 \, \text{m/s} \):
\[ KE = \frac{1}{2} mv_{max2}^2 = \frac{1}{2} m (4)^2 = 8m \]
Using the photoelectric equation again:
\[ E' = \phi + KE \]
Where \( E' = \frac{hc}{\lambda'} \). Thus:
\[ \frac{hc}{\lambda'} = \frac{hc}{3\lambda} + 8m \]

### Step 6: Rearranging the second equation
Rearranging gives:
\[ \frac{hc}{\lambda'} - \frac{hc}{3\lambda} = 8m \]
\[ \frac{hc}{\lambda'} = \frac{hc}{3\lambda} + 8m \]

### Step 7: Substitute \( \frac{hc}{\lambda} \)
From Step 4, we know \( \frac{hc}{\lambda} = 3m \). Substitute this into the equation:
\[ \frac{hc}{\lambda'} = \frac{1}{3}(3m) + 8m = m + 8m = 9m \]

### Step 8: Solve for the new wavelength
Now, we can express \( \lambda' \):
\[ \frac{hc}{\lambda'} = 9m \]
\[ \lambda' = \frac{hc}{9m} \]

### Step 9: Convert to nanometers
To convert \( \lambda' \) into nanometers, we can use the known values of \( h \) and \( c \) (Planck's constant and speed of light). However, since the problem does not specify numerical values for \( h \) and \( c \), we can leave it in terms of \( hc \).

### Final Answer
The new wavelength \( \lambda' \) is given by:
\[ \lambda' = \frac{hc}{9m} \]