Question

The total energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving the spectral line of lowest energy in the visible region of its atomic spectrum is:
$[\text{Given}:E_{1\left(H\right)}=2.176\times {10}^{-18}]$

• A. $1.82\times {10}^{5}J/mol$
• B. $1.96\times {10}^{5}J/mol$
• C. $2.01\times {10}^{5}J/mol$
• D. $2.06\times {10}^{5}J/mol$

Answer 1

The correct option is A $1.82\times {10}^{5}J/mol$

Formula used, $\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$R=1.109678\times {10}^7{m}^{-1}$, for visible region we have Balmer series.

For Balmer series, $n_1=2$ and $n_2=3\left(forlowestenergy\right)$
Substituting values,

$\frac{1}{\lambda }=1.109678\times {10}^7[\frac{1}{2^2}-\frac{1}{3^2}]$

$\Rightarrow \lambda =6.546\times {10}^{-7}m$

For energy formula is,

$\Delta E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{6.546\times {10}^{-7}}=3.037\times {10}^{-19}J$

Now energy emitted from 1.0 g of hydrogen atom $\left(E_H\right)=3.037\times {10}^{-19}\times N_A=3.037\times {10}^{-19}\times 6.02\times {10}^{23}=1.82\times {10}^{5}J$

Hence, the correct option is $A$