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$-$1]
Answer (integer)
181
Solution
<p>Energy of EMR = IE of the metal (A)</p>
<p>$= hv = {{hc} \over \lambda }$ atom<sup>$-$1</sup> $= {{hc} \over \lambda } \times {N_A}$ mol<sup>$-$1</sup></p>
<p>$$ = {{(6.63 \times {{10}^{ - 34}}) \times (3 \times {{10}^8}) \times (6.02 \times {{10}^{23}})} \over {(663 \times {{10}^{ - 9}})}}$$ J mol<sup>$-$1</sup> [$\because$ $\lambda$ = 663 nm = 663 $\times$ 10<sup>$-$9</sup> m]</p>
<p>= 180600 J mol<sup>$-$1</sup> = 180.6 kJ mol<sup>$-$1</sup> $\simeq$ 181 kJ mol<sup>$-$1</sup></p>
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
This question is part of PrepWiser's free JEE Main question bank. 122 more solved questions on Atomic Structure are available — start with the harder ones if your accuracy is >70%.