Question

If the energy of incident radiation is doubled, how will the maximum kinetic energy of the photoelectrons change?

• A. becomes double
• B. becomes more than double
• C. becomes half
• D. doesn't change

Answer 1

The correct option is B becomes more than double

Let say
$(K.E{)}_{max1}=E-{\varphi }_0.....(1)$
If the energy of incident radiation is doubled,
$(K.E{)}_{max2}=2E-{\varphi }_0$
$=E+(E-{\varphi }_0)$
$K.E_{max2}=E+K.E_{max1}$ ---from $\left(1\right)$
As, $E>K.E_{max1}$
$\Rightarrow K.E_{max2}>2K.E_{max1}$

Hence, $\left(B\right)$ is the correct answer.

Why this question? The question tests thorough understanding of the formula, corresponding manipulation and predictions of the same.