Question

When a metallic surface is illuminated with radiation of wavelength $\lambda$ , the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2 \lambda$, the stopping potential is $\frac{V}{4}$ . The threshold wavelength for the metallic surface is :

• A. $5 \lambda$
• B. $\frac{5}{2} \lambda$
• C. $ 3 \lambda$
• D. $4 \lambda$

Answer 1

The Correct Option is C

To find the threshold wavelength for a metallic surface, we need to use the concept of the photoelectric effect. The energy of a photon is related to its wavelength by the equation:

E = \frac{hc}{\lambda}

where h is Planck's constant and c is the speed of light. The maximum kinetic energy of the emitted electrons is given by:

K.E. = eV = \frac{hc}{\lambda} - \phi

where \phi is the work function of the metal and V is the stopping potential.

1. When the metallic surface is illuminated with radiation of wavelength \lambda, the stopping potential is V. Therefore, the equation becomes:

eV = \frac{hc}{\lambda} - \phi \quad \text{(1)}

1. When the same surface is illuminated with radiation of wavelength 2\lambda, the stopping potential is \frac{V}{4}. Therefore, the equation becomes:

e\left( \frac{V}{4} \right) = \frac{hc}{2\lambda} - \phi \quad \text{(2)}

1. Subtract equation (2) from equation (1) to eliminate \phi:

eV - e \left( \frac{V}{4} \right) = \frac{hc}{\lambda} - \frac{hc}{2\lambda}

\Rightarrow \frac{3eV}{4} = \frac{hc}{2\lambda}

\Rightarrow \frac{eV}{2} = \frac{hc}{3\lambda}

1. The threshold wavelength \lambda_0 is when kinetic energy is zero, hence eV = 0 in the above equation:

\phi = \frac{hc}{\lambda_0}

1. From equation (1), inserting eV = 0, we get:

0 = \frac{hc}{\lambda_0} - \phi

eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}

1. Thus, from our derived expression \phi = \frac{hc}{3\lambda}, equating with \phi = \frac{hc}{\lambda_0},

\lambda_0 = 3\lambda

Therefore, the threshold wavelength for the metallic surface is 3 \lambda.