An electron changes its position from orbit n | vedclass

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Question

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

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$\frac{16}{3R}$

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(B) The Rydberg formula for the wavelength $\lambda$ of emitted radiation is given by:$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$Here,the electron transitions from $n_2 = 4$ to $n_1 = 2$.Substituting the values:$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$$\frac{1}{\lambda} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$$\frac{1}{\lambda} = R \left[ \frac{4 - 1}{16} \right] = R \left[ \frac{3}{16} \right]$Therefore,$\lambda = \frac{16}{3R}$.

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 $n_1=2$	$IV$. $486.1$

An electron changes its position from orbit $n = 4$ to the orbit $n = 2$ of an atom. The wavelength of the emitted radiation is ($R =$ Rydberg's constant).

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