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$-$1)
Solution
<p>Energy of one photon</p>
<p>$E = {{1240} \over {\lambda (nm)}}eV$</p>
<p>$= {{1240} \over {300}}$</p>
<p>$= 4.1333\,eV$</p>
<p>$\therefore$ Energy of one mole of photon</p>
<p>$= 4.1333 \times 6.02 \times {10^{23}}\,eV$</p>
<p>$= 4.1333 \times 6.02 \times {10^{23}} \times 1.6 \times {10^{ - 19}}\,J$</p>
<p>$$ = {{4.1333 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}\,kJ$$</p>
<p>$= 399\,kJ/mol$</p>
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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