Verified Solution • 2025–26 Session moderate

QJEEM21Ch924d: JEE Main 2021 Chemistry - Class 11 JEEM21Ch924d Solution: Electromagnetic radiation of wavelength 663 nm is just…

Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization enegy of metal A in kJ mol −- 1 is __________. (Rounded off to the nearest integer)

[h = 6.63 ×\times 10 −- 34 Js, c = 3.00 ×\times 108 ms −- 1, NA = 6.02 ×\times 1023 mol −- 1]

Step-by-Step Verified Solution

Energy of EMR = IE of the metal (A)

=hv=hcλ= hv = {{hc} \over \lambda } atom −- 1 =hcλ×NA= {{hc} \over \lambda } \times {N_A} mol −- 1

=(6.63×10−34)×(3×108)×(6.02×1023)(663×10−9)= {{(6.63 \times {{10}^{ - 34}}) \times (3 \times {{10}^8}) \times (6.02 \times {{10}^{23}})} \over {(663 \times {{10}^{ - 9}})}} J mol −- 1 [ ∵\because λ\lambda = 663 nm = 663 ×\times 10 −- 9 m]

= 180600 J mol −- 1 = 180.6 kJ mol −- 1 ≃\simeq 181 kJ mol −- 1

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