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The shortest wavelength of H atom in the Lyman series is $\lambda_{1}$ . The longest wavelength in the Balmer series He⁺ is:
Option: 1 $\frac{36 \lambda _{1}}{5}$
Option: 2 $\frac{5\lambda _{1}}{9}$
Option: 3 $\frac{9 \lambda _{1}}{5}$
Option: 4 $\frac{27 \lambda _{1}}{5}$

Answers (1)

best_answer

We know,

$\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$

So, $\lambda=\frac{h c}{\Delta E}$ for $\lambda$ minimum(Shortest)

$\mathrm{\Delta \mathrm{E}= maximum}$

for f H atoms in the Lyman series n = 1 & for $\mathrm{\Delta \mathrm{E}_{maximum}}$ Transition must be form n = $\infty$ to n = 1,

Z = 1

$\text { So } \frac{1}{\lambda}=R_{H} Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}} \mathrm{(1)}^{2}(1-0)$

$\frac{1}{\lambda_1}=\mathrm{R_H} \times(1)^{2} \Rightarrow \lambda_{1}=\frac{1}{\mathrm{R_H}}$

For longest wavelength $\mathrm{\Delta \mathrm{E}_{minimum}}$ for Balmer series n = 3 to n = 2 will have $\mathrm{\Delta \mathrm{E}_{minimum}}$ for He⁺ , Z = 2

$\text { So } \frac{1}{\lambda_{2}}=R_{H} \times Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{R}_{\mathrm{H}} \times (2)^2\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{R}_{\mathrm{H}} \times \frac{5}{9}$

$\lambda_{2}=\frac{1}{R_H} \times \frac{9}{5}$

$\lambda_{2}=\lambda_{1} \times \frac{9}{5}$

Therefore, the correct option is (3).

Posted by

Kuldeep Maurya

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