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Question: Answered & Verified by Expert
The work functions of $\mathrm{Ag}, \mathrm{Mg}, \mathrm{K}$ and $\mathrm{Na}$ respectively in $\mathrm{eV}$ are $4.3,3.7,2.25,2.30$. When an electromagnetic radiation of wavelength of $300 \mathrm{~nm}$ is allowed to fall on these metal surface, the number of metals from which the electrons are ejected is
$$
\left(\mathbf{e V}=1.6022 \times 10^{-19} \mathrm{~J}\right)
$$
ChemistryStructure of AtomAP EAMCETAP EAMCET 2017 (26 Apr Shift 1)
Options:
  • A 4
  • B 3
  • C 2
  • D 5
Solution:
2852 Upvotes Verified Answer
The correct answer is: 3
Energy of EM wave used
$$
\begin{aligned}
& E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \\
& E=0.066 \times 10^{-17}=6.6 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
Metals surface which have work function less than or equal to $6.6 \times 10^{-19} \mathrm{~J}$ will eject electrons on radiation with light of $300 \mathrm{~nm}$.
Work function of $\mathrm{Ag}=4.3 \mathrm{eV} \times 1.6022 \times 10^{-19}=6.89 \times$ $10^{-19} \mathrm{~J}$
$$
\begin{aligned}
& \mathrm{Mg}=3.7 \mathrm{eV} \times 1.6022 \times 10^{-19}=5.93 \times 10^{-19} \mathrm{~J} \\
& \mathrm{~K}=2.25 \mathrm{eV} \times 1.6022 \times 10^{-19}=3.60 \times 10^{-19} \mathrm{~J} \\
& \mathrm{Na}=2.30 \mathrm{eV} \times 1.6022 \times 10^{-19}=3.68 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
$\therefore \quad$ Only $\mathrm{Mg}, \mathrm{K}$ and $\mathrm{Na}$ have their work function less than $6.6 \times 10^{-19} \mathrm{~J}$ hence, they will eject the electrons.

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