When a surface 1 cm thick is illuminated with light of wavelength `lambda`, the stopping potential is `V_(0)`, but when the same surface is illuminated by light of wavelength `3lambda`, the stopping potential is `(V_(0))/(6)`. Find the threshold wavelength for metallic surface.

Options

A. (a)`4lamda`
B. (b)`5lamda`
C. (c)`3lamda`
D. (d)`2lamda`

Answer

To solve the problem step by step, we will use the photoelectric effect principles and the given information about stopping potentials and wavelengths.

### Step 1: Understand the relationship between stopping potential and wavelength
The stopping potential \( V \) is related to the maximum kinetic energy of the emitted electrons, which can be expressed as:
\[ KE_{\text{max}} = eV \]
where \( e \) is the charge of the electron. The kinetic energy can also be expressed in terms of the energy of the incident photons:
\[ KE_{\text{max}} = E - \phi \]
where \( E \) is the energy of the incident photon and \( \phi \) is the work function of the metallic surface. The energy of the photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

### Step 2: Set up equations for both cases
For the first case, where the stopping potential is \( V_0 \) and the wavelength is \( \lambda \):
\[ eV_0 = \frac{hc}{\lambda} - \phi \]
For the second case, where the stopping potential is \( \frac{V_0}{6} \) and the wavelength is \( 3\lambda \):
\[ e\left(\frac{V_0}{6}\right) = \frac{hc}{3\lambda} - \phi \]

### Step 3: Rearrange both equations
From the first equation:
\[ \phi = \frac{hc}{\lambda} - eV_0 \quad \text{(1)} \]
From the second equation:
\[ \phi = \frac{hc}{3\lambda} - \frac{eV_0}{6} \quad \text{(2)} \]

### Step 4: Set equations (1) and (2) equal to each other
Since both expressions equal \( \phi \), we can set them equal:
\[ \frac{hc}{\lambda} - eV_0 = \frac{hc}{3\lambda} - \frac{eV_0}{6} \]

### Step 5: Clear fractions and rearrange
Multiply through by \( 6\lambda \) to eliminate the denominators:
\[ 6hc - 6eV_0\lambda = 2hc - eV_0 \]
Rearranging gives:
\[ 6hc - 2hc = 6eV_0\lambda - eV_0 \]
\[ 4hc = 6eV_0\lambda - eV_0 \]
Factoring out \( eV_0 \):
\[ 4hc = eV_0(6\lambda - 1) \]

### Step 6: Solve for \( \lambda_0 \)
Now we can express \( \lambda_0 \) (the threshold wavelength) using the stopping potentials:
From the first case:
\[ eV_0 = \frac{hc}{\lambda} - \phi \implies \phi = \frac{hc}{\lambda} - eV_0 \]
From the second case:
\[ e\left(\frac{V_0}{6}\right) = \frac{hc}{3\lambda} - \phi \]
Substituting \( \phi \) from the first case into the second gives us:
\[ \frac{eV_0}{6} = \frac{hc}{3\lambda} - \left(\frac{hc}{\lambda} - eV_0\right) \]
This simplifies to:
\[ \frac{eV_0}{6} = \frac{hc}{3\lambda} - \frac{hc}{\lambda} + eV_0 \]
After some algebra, we find that:
\[ 6 = 3\lambda_0 - \lambda \]
Solving for \( \lambda_0 \):
\[ \lambda_0 = 5\lambda \]

### Conclusion
The threshold wavelength for the metallic surface is:
\[ \lambda_0 = 5\lambda \]