In a hydrogen-like atom,when an electron tran | vedclass

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Question

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Vedclass · Accepted Answer

$486$

Vedclass · Answer

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Since the atom is the same,$RZ^2$ is constant,so $\frac{1}{\lambda} \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first transition ($n=5$ to $n=2$): $\frac{1}{434} = k \left( \frac{1}{2^2} - \frac{1}{5^2} ight) = k \left( \frac{1}{4} - \frac{1}{25} ight) = k \left( \frac{21}{100} ight)$.For the second transition ($n=4$ to $n=2$): $\frac{1}{x} = k \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = k \left( \frac{1}{4} - \frac{1}{16} ight) = k \left( \frac{3}{16} ight)$.Dividing the two equations: $\frac{x}{434} = \frac{21/100}{3/16} = \frac{21}{100} 	imes \frac{16}{3} = \frac{7 	imes 16}{100} = \frac{112}{100} = 1.12$.Therefore,$x = 434 	imes 1.12 = 486.08 \ nm \approx 486 \ nm$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

In a hydrogen-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \ nm$ is emitted. What will be the wavelength (in $nm$) of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

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