Question

The photoelectric threshold wavelength of a certain metal is $3315A.V.$ Its work function is....

• A. $6\times {10}^{-19}J$
• B. $7.286\times {10}^{-19}J$
• C. $9\times {10}^{-19}J$
• D. $9.945\times {10}^{-19}J$

Answer 1

The correct option is A $6\times {10}^{-19}J$

The work function of the metal

$\varphi =\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{3315\times {10}^{-10}}=5.996\times {10}^{-19}J\approx 6\times {10}^{-19}$

here, $\lambda =Thresholdwavelength$

Hence, Its work function is $6\times {10}^{-19}J$