Question

When photons of energy hv fall on an aluminum plate (of work function E₀), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be :

• A. K + E₀
• B. 2K
• C. K
• D. k + hv

Answer 1

The Correct Option is D

To solve this problem, we must understand the concept of the photoelectric effect. According to Einstein's photoelectric equation, the energy of the incident photon is used to overcome the work function of the metal and provide kinetic energy to the ejected photoelectrons.

The equation can be written as:

hv = E_0 + K

Here,

• hv is the energy of the incident photon.
• E_0 is the work function of the metal.
• K is the maximum kinetic energy of the photoelectrons.

From the equation, the maximum kinetic energy can be expressed as:

K = hv - E_0

When the frequency of the radiation is doubled, the new frequency becomes 2v. Thus, the energy of the incident photons becomes 2hv.

Using the photoelectric equation for the doubled frequency, we have:

2hv = E_0 + K_{\text{new}}

Solving for the new maximum kinetic energy K_{\text{new}}:

K_{\text{new}} = 2hv - E_0

Substitute the original expression for K into the equation:

K_{\text{new}} = 2hv - (hv - K)

Simplifying this, we get:

K_{\text{new}} = K + hv

Therefore, with the frequency doubled, the maximum kinetic energy of the ejected photoelectrons increases and becomes K + hv.

Conclusion: The correct answer is K + hv.