Question

The work function for tungsten and sodium are $4.5 \,eV$ and $2.3 \,eV$ respectively. If the threshold wavelength $\lambda$ for sodium is $5460\, \mathring{A}$, the value of $\lambda$ for tungsten is

• A. $528 \mathring{A}$
• B. $2791 \mathring{A}$
• C. $5893\mathring{A}$
• D. $10683 \mathring{A}$

Answer 1

Answer: (B)

Since, $\phi_{0}=\frac{h c}{\lambda_{0}}$;
so $\frac{\phi_{0_{r}}}{\phi_{0_{\text {Nu }}}}=\frac{\lambda_{ Na }}{\lambda_{ T }}$
or $\lambda_{ T }=\lambda_{ Na } \times \frac{\phi_{0_{ N }}}{\phi_{0_{ T }}}$
$=\frac{5460 \times 2.3}{4.5}=2791\, \mathring{A}$