Question

Threshold frequencies of the metals \(A\) and \(B\) are respectively \[ 4\times10^{14}\ \text{Hz} \] and \[ 6\times10^{14}\ \text{Hz}. \] If both are irradiated with light of frequency \[ 10^{15}\ \text{Hz}, \] what is the ratio of the kinetic energies of electrons emitted from \(A\) and \(B\)?

• A. \(2:3\)
• B. \(3:2\)
• C. \(4:9\)
• D. \(9:4\)

Hint:

For photoelectric emission, \[ K_{\max}=h(\nu-\nu_0). \] For the same incident frequency, the metal with the lower threshold frequency emits photoelectrons with greater kinetic energy.

Answer 1

The Correct Option is B

Concept: According to Einstein's photoelectric equation, \[ K_{\max}=h(\nu-\nu_0), \] where \[ \nu=\text{frequency of incident light}, \] \[ \nu_0=\text{threshold frequency}. \]

Step 1: Find the kinetic energy for metal \(A\). Given, \[ \nu=10^{15}\ \text{Hz}, \qquad \nu_{0A}=4\times10^{14}\ \text{Hz}. \] Hence, \[ K_A = h\left(10^{15}-4\times10^{14}\right). \] \[ = h(6\times10^{14}). \]

Step 2: Find the kinetic energy for metal \(B\). Given, \[ \nu_{0B}=6\times10^{14}\ \text{Hz}. \] Therefore, \[ K_B = h\left(10^{15}-6\times10^{14}\right). \] \[ = h(4\times10^{14}). \]

Step 3: Calculate the ratio. \[ \frac{K_A}{K_B} = \frac{h(6\times10^{14})} {h(4\times10^{14})}. \] \[ = \frac{6}{4}. \] \[ = \frac{3}{2}. \] Therefore, \[ \boxed{K_A:K_B=3:2} \] \[ \boxed{\text{Answer = (B)}} \]