Question

Which of the following expressions represents the spectrum of Balmer series (If n is the
principal quantum number of higher energy level ) in a
hydrogen atom

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Answer 1

answer : option (B) $\boxed{\bar{v}=\frac{R(n-2)(n+2)}{4n^2}}$

explanation : we know in case of Balmer's series , transition start from 2nd energy level. so, $n_1=2$ to $n_2$= n

from Rydberg's equation ,

$\bar{v}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where R is Rydberg's constant, Z is atomic number.

for hydrogen - atom, Z = 1

so, $\bar{v}$ = R(1)² [ 1/2² - 1/n²]

= R[ (n² - 2²)/2²n² ]

= R(n - 2)(n + 2)/4n²

hence, $\boxed{\bar{v}=\frac{R(n-2)(n+2)}{4n^2}}$

so, option (B) is correct choice.