Question

If work function of sodium $\left(\text{Na}\right)$ metal is $2.75\text{eV}$, its threshold wavelength lies in

• A. Visible region
• B. Infrared region
• C. Radio region
• D. Ultraviolet region

Answer 1

The correct option is A Visible region

Work function ${\varphi }_0=\frac{hc}{{\lambda }_0}$

${\lambda }_0=\frac{hc}{{\varphi }_0}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{2.75\times 1.6\times {10}^{-19}}=452\text{nm}$

Now, $452\text{nm}$ falls into region of visible light $(400\text{nm}-700\text{nm})$

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

Why this question? This question tests the combined knowledge of photoelectric effect & Electromagnetic waves.