Electromagnetic radiation of wavelength

Question:

Electromagnetic radiation of wavelength $663 \mathrm{~nm}$ is just sufficient to ionize the atom of metal $\mathrm{A}$.

The ionization energy of metal $\mathrm{A}$ in $\mathrm{kJmol}^{-1}$ is _____________. (Rounded off to the nearest integer)

$\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3.00 \times 10^{8} \mathrm{~ms}^{-1}, \mathrm{~N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right]$

Solution:

(180)

Whergy req. to ionize an atom of metal ' $A^{\prime}=\frac{h c}{\lambda}=\frac{h c}{663 \mathrm{~nm}}$

for 1 mole atoms of ' A $^{\prime}$ Total energy required $=N_{A} \times \frac{h c}{\lambda}$

$=\frac{6.023 \times 10^{23} \times 6.63 \times 10^{-34} \times 3 \times 10^{8}}{663 \times 10^{-9}}$

$=6.023 \times 3 \times 10^{23-34+8+7}$

$=18.04 \times 10^{4} \mathrm{~J} / \mathrm{mol}$

$=180.4 \mathrm{KJ} / \mathrm{mol}$

Nearest Integer $=180 \mathrm{KJ} / \mathrm{Mol}$

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