Question

The following data is observed in the photoelectric effect experiment when cesium was usedas metal. The threshold frequency of the metal is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$h{v}_0$ = $hv-\frac{1}{2}m{v}^2$

$v_0=\frac{\frac{1}{2}m{v}^2}{h}$

= $\frac{3\times {10}^8}{5\times {10}^7}-\frac{1}{2}\frac{9\times {10}^{-31}\times (4.87\times {10}^5{)}^2}{6.625\times {10}^{-34}}$
= $6\times {10}^{14}-16.1\times {10}^{13}$

= $4.4\times {10}^{14}se{c}^{-1}$