frac{x}{x+4} is the ratio of energies of phot | vedclass

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(B) The energy of a photon emitted during a transition from level $n_2$ to $n_1$ is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right

Vedclass · Accepted Answer

$5$

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(B) The energy of a photon emitted during a transition from level $n_2$ to $n_1$ is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.For transition $(i)$ from $n=3$ to $n=2$: $E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight)$.For transition $(ii)$ from $n=\infty$ to $n=2$: $E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = 13.6 \left( \frac{1}{4} - 0 ight) = 13.6 \left( \frac{1}{4} ight)$.The ratio is $\frac{E_1}{E_2} = \frac{13.6(5/36)}{13.6(1/4)} = \frac{5/36}{1/4} = \frac{5}{9}$.Given the ratio is $\frac{x}{x+4}$,we have $\frac{x}{x+4} = \frac{5}{9}$.Cross-multiplying gives $9x = 5x + 20$,which simplifies to $4x = 20$,so $x = 5$.

$\frac{x}{x+4}$ is the ratio of energies of photons produced due to the transition of an electron of a hydrogen atom from its $(i)$ third permitted energy level to the second level and $(ii)$ the highest permitted energy level to the second permitted level. The value of $x$ will be.

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