JEE PYQ: Atomic Structure Question 12
Question 12 - 2021 (25 Feb Shift 2)
Electromagnetic radiation of wavelength 663 nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol$^{-1}$ is ____.
[$h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^{8}$ m s$^{-1}$, $N_A = 6.02 \times 10^{23}$ mol$^{-1}$]
Show Answer
Answer: 180
Solution
$IE = N_A \times \frac{hc}{\lambda} = \frac{6.023 \times 10^{23} \times 6.63 \times 10^{-34} \times 3 \times 10^8}{663 \times 10^{-9}} = 180.4$ kJ/mol $\approx 180$.
BETA
