Question

The photoelectric threshold frequency of a material is $\nu$.When the light of frequency $4\nu$ is incident on the metal, the maximum kinetic energy of the emitted photoelectron is?

• A. $3h\nu$
• B. $4h\nu$
• C. $5h\nu$
• D. $6h\nu$

Answer 1

The correct option is A

$3h\nu$

Step 1: Given data

${\nu }_1=4\nu$

$v_0=\nu$

Here,

v is the photoelectric threshold frequency.

4v is the frequency of light.

Step 2: To find

We have to determine the maximum kinetic energy of the emitted photoelectron.

Step 3: Calculate the kinetic energy

The threshold frequency is the lowest frequency of incident light at which electron emission begins. It varies depending on the material.

The work function is the minimum energy required to initiate the photoelectric effect from the metal surface.

$$\begin{gathered} {\text{K.E}}_{MAX}\text{=}h{\nu }_1-h{\nu }_0 \\ =h\left(4\nu \right)-h\nu \\ =3h\nu \end{gathered}$$

Hence, option(A) is correct.