Question

If the work function of a photometal is 6.875eV, its threshold wavelength will be $(Takec=3\times {10}^{8}m/s)$

• A. 3600 $\mu =0.3$
• B. 2400 $\therefore$
• C. 1800 $F=0.3\times 250=75N.$
• D. 1200 $\overrightarrow{\mathbf{F}}$

Answer 1

Answer: (C)

The minimum energy required for the emission of photoelectron from a metal is called work function of that metal, $W=\frac{hc}{\lambda }$ where h is Plancks constant, c is speed of light, $\lambda$ is wavelength. $\Rightarrow$ $\lambda =\frac{hc}{W}$ $=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{6.825\times 1.6\times {10}^{-19}}$ $=1.8\times {10}^{-7}$ $=1800\times {10}^{-10}m$ $=1800\stackrel{\text{o}}{A}$