Question

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

• A. $E_n=\frac{-21.8\times 10^{-19}}{n^2}Jato{m}^{-1}$
• B. $E_n=\frac{-13.6}{n^2}eVato{m}^{-1}$
• C. $E_n=\frac{-1312}{n^2}kJmo{l}^{-1}$
• D. $E_n=\frac{-12.8\times 10^{-19}}{n^2}ergato{m}^{-1}$

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 $1ev=1.602\times 10^{-19}J$ 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 $kJmo{l}^{-1}$ is
$E_n=\frac{-21.8\times 10^{-19}}{1000h^2}\times 6.022\times 10^{23}kJ/mol$
$E_n=\frac{-1312}{n^2}kJ/mol$ 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