JEE MAIN - Chemistry (2022 - 24th June Evening Shift - No. 2)
The energy of one mole of photons of radiation of wavelength 300 nm is
(Given : h = 6.63 $$\times$$ 10$$-$$34 J s, NA = 6.02 $$\times$$ 1023 mol$$-$$1, c = 3 $$\times$$ 108 m s$$-$$1)
(Given : h = 6.63 $$\times$$ 10$$-$$34 J s, NA = 6.02 $$\times$$ 1023 mol$$-$$1, c = 3 $$\times$$ 108 m s$$-$$1)
235 kJ mol$$-$$1
325 kJ mol$$-$$1
399 kJ mol$$-$$1
435 kJ mol$$-$$1
Explanation
Energy of one photon
$$E = {{1240} \over {\lambda (nm)}}eV$$
$$ = {{1240} \over {300}}$$
$$ = 4.1333\,eV$$
$$\therefore$$ Energy of one mole of photon
$$ = 4.1333 \times 6.02 \times {10^{23}}\,eV$$
$$ = 4.1333 \times 6.02 \times {10^{23}} \times 1.6 \times {10^{ - 19}}\,J$$
$$ = {{4.1333 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}\,kJ$$
$$ = 399\,kJ/mol$$
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