Question

If the maximum kinetic energy of an emitted electron is $4$ times the work function of the metal $\text{Li}$, what is the wavelength of the incident radiation?
Given: work function of $\text{Li}=2.5\text{eV}$

• A. $995\mathring{\text{A}}$
• B. $9950\text{pm}$
• C. $995\text{nm}$
• D. $995\mu m$

Answer 1

The correct option is A $995\mathring{\text{A}}$

According to the equation of photoelectric effect,
$(K.E{)}_{max}=\frac{hC}{\lambda }-{\varphi }_0......(1)$

As, $(K.E{)}_{max}=4\times {\varphi }_0$

$\Rightarrow 4{\varphi }_0=\frac{hC}{\lambda }-{\varphi }_0$

$\Rightarrow 5{\varphi }_0=\frac{hC}{\lambda }$

$5\times 2.5\times 1.6\times {10}^{-19}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{\lambda }$

$\therefore \lambda =994.5\mathring{\text{A}}\approx 995\mathring{\text{A}}$

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

Why this question? This equation is a typical question & can be asked in NEET directly. Concept : $(K.E{)}_{max}=\frac{hC}{\lambda }=\varphi$