Electromagnetic radiation of... | Atomic Structure JEE Main 2021

JEE MAIN 2021

25-02-2021 S2

Question

Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A . The ionization energy of metal $A$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is $\_\_\_\_$ (Rounded-off to the nearest integer) $$ \begin{aligned} & {\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3.00 \times 10^8 \mathrm{~ms}^{-1}\right.} \\ & \left.\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right] \end{aligned} $$

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Solution

$\begin{aligned} & E=\frac{h c}{\lambda} \times \frac{N_A}{1000} \\ & =\frac{6.63 \times 10^{-34} \times 3 \times 10^8 \times 6.02 \times 10^{23}}{663 \times 10^{-9} \times 1000} \\ & =3 \times 6.02 \times 10 \mathrm{~kJ} \\ & =180.6 \mathrm{~kJ}\end{aligned}$

Question Tags

JEE Main

Chemistry

Easy

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