Question

The work functions of $Ag,Mg,K$ and $Na$ respectively in $eV$ are $4.3,3.7,2.25,2.30$ When an electromagnetic radiation of wavelength of $300nm$ is allowed to fall on these metal surface, the number of metals from which the electrons are ejected is $\left(1eV=1.6022\times 10^{-19}J\right)$

• A. 4
• B. 3
• C. 2
• D. 5

Answer 1

Answer: (B)

Work function for $Ag,Mg,K$ and Na in joules are for

$Ag=4.3eV\times 1.6022\times 10^{-19}=6.89\times 10^{-19}J$

$Mg=3.7eV\times 1.6022\times 10^{-19}=5.93\times 10^{-19}J$

$K=2.25eV\times 1.6022\times 10^{-19}=3.60\times 10^{-19}J$

$Na=2.30eV+1.6022\times 10^{-19}=3.68\times 10^{-19}J$

Energy used is of wave length $(\lambda =300nm)$.

$E_{300}=\frac{hC}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{300\times 10^{-9}}$

$E_{300}=\frac{19.80\times 10^{-26}}{300\times 10^{-9}}=0.066\times 10^{-17}$

$=6.6\times 10^{-19}J$

$\because$ Metal will eject electrons only when $W\le E_{300}$

$\therefore$ Only $Mg,K$ and Na (3-metals) will eject the electrons.