Question

Energy of incident radiation increases from $5\text{eV}$ to $10\text{eV}$. If the threshold wavelength for the given metal surface is $310\text{nm}$, find the ratio of maximum kinetic energies of the emitted electrons.

• A. $1:6$
• B. $1:3$
• C. $1:5$
• D. $1:2$

Answer 1

The correct option is A $1:6$

${\varphi }_0=\text{work function}=\frac{hc}{\lambda }$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{310\times {10}^{-9}}=4\text{eV}$

Now, $K.E_{max}=E-{\varphi }_0$
As the material is same, ${\varphi }_0=4\text{eV}$
$\Rightarrow \frac{(K.E{)}_{max1}}{(K.E{)}_{max2}}=\frac{E_1-{\varphi }_0}{E_2-{\varphi }_0}=\frac{5-4}{10-4}=\frac{1}{6}$

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

$\begin{array}{c}\text{Why this question?}\\ \text{This question shows dependance of maximum kinetic energy on energy of incident radiation.}\end{array}$