Question

If threshold frequency for a metal is $5\times {10}^{14}\text{Hz}$. Find the ratio of $(K.E{)}_{max}$ of emitted electrons when its surface is irradiated with radiations of frequencies $7\times {10}^{14}\text{Hz}$ and $9\times {10}^{14}\text{Hz}$ respectively.

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

Answer 1

The correct option is A $1:2$

$(K.E{)}_{max}=E-{\varphi }_0=hf-h{f}_0=h(f-f_0)$

$\therefore K.E_{max}\propto (f-f_0)$

as $h\to$ planks constant

$\therefore \frac{(K.E{)}_{max1}}{(K.E{)}_{max2}}=\frac{(7-5)}{(9-5)}\times \frac{{10}^{14}}{{10}^{14}}=\frac{1}{2}$

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

Why this question? Concept: $K.E_{max}\propto (f-f_0)$ where $f\to$ frequency of incident radiations and $f_0=$ threshold frequency.