Question

The Bohr’s energy of a stationary state of hydrogen atom is given as $E_{n}=\frac{-2\pi^{2}\,me^{4}}{n^{2}\,h^{2}}$. Putting the values of $m$ and $e$ for $n^{th}$ energy level which is not the correct value?

• A. $E_{n}=\frac{-21.8 \times 10^{-19}}{n^{2}}\,J\,atom^{-1}$
• B. $E_{n}=\frac{-13.6}{n^{2}}\,eV\,atom^{-1}$
• C. $E_{n}=\frac{-1312}{n^{2}}\,kJ\,mol^{-1}$
• D. $E_{n}=\frac{-12.8 \times 10^{-19}}{n^{2}}\,erg\,atom^{-1}$

Answer 1

Answer: (D)

given $E_{n}=\frac{-2 \pi m e^{4}}{n^{2} h^{2}}$ on putting the values of $m$ and $e$ we get the value in ev which is given as,
$E_{n}=\frac{13.6}{n^{2}}$ ev/atom (i)
Hence option $B$ of represent correct value.
since $1\, ev =1.602 \times 10^{-19} J$ put in 1
$E_{n}=\frac{-21.8 \times 10^{-19}} {n^2}J$ atom (ii)
hence option (A) also represent correct value. for option (C) we know that mole Contains $6.022 \times 10^{23}$ atom so value of energy in $kJ mol^{-1}$ is
$E_{n}=\frac{-21.8 \times 10^{-19}}{1000 h ^{2}} \times 6.022 \times 10^{23} kJ / mol$
$E_{n}=\frac{-1312}{n^{2}} kJ /mol$ option ' $C$ ' also represent correct value.
for option (D) we know that lerg $=10^{-7} J$ wing this in eq(ii) we get
$E_{n}=\frac{-21.8 \times 10^{-12}}{n^{2}} erg$ of so option D represent incorrect value