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 $300 nm$ is allowed to fall on these metal surface, the number of metals from which the electrons are ejected is $\left(1\, eV =1.6022 \times 10^{-19}\, J \right)$

• A. 4
• B. 3
• C. 2
• D. 5

Answer 1

Answer: (B)

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

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

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

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

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

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

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

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

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

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

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