The electron in hydrogen atom is in the first Bohr orbit (n=1 ). The ratio of transition energies, E(n=1 arrow n =3 ) to E(n=1 arrow n=2 ), is-

Correct answer: (A) 32 / 27. Δ E=13.6 (frac1n_1^2-frac1n_2^2 ) eV / atom Δ E_1 arrow 3=fracfrac11^2-frac13^2frac11^2-frac12^2=.32/27 Δ E_1 arrow 2

Q26 / 90 · KVPY 2011 (SB/SX) · Chemistry

The electron in hydrogen atom is in the first Bohr orbit $(\mathrm{n}=1$ ). The ratio of transition energies, $\mathrm{E}(\mathrm{n}=1 \rightarrow \mathrm{n}$ $=3$ ) to $\mathrm{E}(\mathrm{n}=1 \rightarrow \mathrm{n}=2$ ), is-

Correct answer
A
$32 / 27$
B
$16 / 27$
C
$32 / 9$
D
$8 / 9$

Solution

$\Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right) \mathrm{eV} / \mathrm{atom}$ $\Delta \mathrm{E}_{1 \rightarrow 3}=\frac{\frac{1}{1^{2}}-\frac{1}{3^{2}}}{\frac{1}{1^{2}}-\frac{1}{2^{2}}}=\frac{.32}{27}$ ${\Delta \mathrm{E}_{1 \rightarrow 2}}$

Hence, the correct answer is option: 'A'.

The electron in hydrogen atom is in the first Bohr orbit $(\mathrm{n}=1$ ). The ratio of transition energies, $\mathrm{E}(\mathrm{n}=1 \rightarrow \mathrm{n}$ $=3$ ) to $\mathrm{E}(\mathrm{n}=1 \rightarrow \mathrm{n}=2$ ), is-

Solution

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