JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 18)
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]
[h = 6.63 $$\times$$ 10$$-$$34 Js, c = 3.00 $$\times$$ 108 ms$$-$$1, NA = 6.02 $$\times$$ 1023 mol$$-$$1]
Answer
181
Explanation
Energy of EMR = IE of the metal (A)
$$ = hv = {{hc} \over \lambda }$$ atom$$-$$1 $$ = {{hc} \over \lambda } \times {N_A}$$ mol$$-$$1
$$ = {{(6.63 \times {{10}^{ - 34}}) \times (3 \times {{10}^8}) \times (6.02 \times {{10}^{23}})} \over {(663 \times {{10}^{ - 9}})}}$$ J mol$$-$$1 [$$\because$$ $$\lambda$$ = 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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