Question

The maximum kinetic energy of the emitted photoelectrons from a photosensitive material of work function \( \phi \), when light of frequency 'v' incidents on it is 'E'. If the frequency of the incident light is 3v, the maximum kinetic energy of the emitted photoelectrons is

• A. \( 3E + 2\phi \)
• B. \( 3E - 2\phi \)
• C. \( 2E + 3\phi \)
• D. \( 2E - 3\phi \)

Hint:

Apply Einstein's photoelectric equation \( K_{max} = h\nu - \phi \) for both given scenarios. Use the information from the first scenario to establish a relationship between \( h\nu \), \( E \), and \( \phi \). Then substitute this relationship into the equation for the second scenario to find the new maximum kinetic energy in terms of \( E \) and \( \phi \).

Answer 1

The Correct Option is A

According to Einstein's photoelectric equation, the maximum kinetic energy \( K_{max} \) of the emitted photoelectrons is given by: $$ K_{max} = h\nu - \phi $$ where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the material.
In the first case, the frequency of the incident light is \( \nu \), and the maximum kinetic energy is \( E \).
So, we have: $$ E = h\nu - \phi \quad \cdots (1) $$ In the second case, the frequency of the incident light is \( 3\nu \).
Let the maximum kinetic energy of the emitted photoelectrons in this case be \( E' \).
Then, according to Einstein's photoelectric equation: $$ E' = h(3\nu) - \phi $$ $$ E' = 3h\nu - \phi \quad \cdots (2) $$ We want to express \( E' \) in terms of \( E \) and \( \phi \).
From equation (1), we can express \( h\nu \) as: $$ h\nu = E + \phi $$ Now, substitute this expression for \( h\nu \) into equation (2): $$ E' = 3(E + \phi) - \phi $$ $$ E' = 3E + 3\phi - \phi $$ $$ E' = 3E + 2\phi $$ Thus, if the frequency of the incident light is \( 3\nu \), the maximum kinetic energy of the emitted photoelectrons is \( 3E + 2\phi \).