An electron absorbs a photon of light of frequency 7.312 × 1014 s -1 in the first excited state of hydrogen and jumps from there to n text th excited state. The value of n is

Question

An electron absorbs a photon of light of frequency 7.312 × 1014 s -1 in the first excited state of hydrogen and jumps from there to n text th excited state. The value of n is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

v =7.312 × 1014 s -1 arrow barv=(v/ c )=2437200 m -1 barv in cm -1=24372 cm -1 For Ist excited state n =2 24372=109677[(1/4)-(1/ n 22)] n 2=6, i.e., 5 text th excited state

Q. An electron absorbs a photon of light of frequency $7.312 \times 1 0^{14} s^{- 1}$ in the first excited state of hydrogen and jumps from there to $n^{th}$ excited state. The value of $n$ is

$v = 7.312 \times 1 0^{14} s^{- 1} \to \overset{v}{ˉ} = \frac{v}{c} = 2437200 m^{- 1}$
$\overset{v}{ˉ}$ in $c m^{- 1} = 24372 c m^{- 1}$
For Ist excited state $n = 2$
$24372 = 109677 [\frac{1}{4} - \frac{1}{n _{2}^{2}}]$
$n_{2} = 6$ , i.e., $5^{th}$ excited state

Q. An electron absorbs a photon of light of frequency 7.312×1014s−1 in the first excited state of hydrogen and jumps from there to nth excited state. The value of n is

v=7.312×1014s−1→vˉ=cv​=2437200m−1 vˉ in cm−1=24372cm−1 For Ist excited state n=2 24372=109677[41​−n22​1​] n2​=6, i.e., 5th excited state

Q. An electron absorbs a photon of light of frequency $7.312 \times 1 0^{14} s^{- 1}$ in the first excited state of hydrogen and jumps from there to $n^{th}$ excited state. The value of $n$ is

$v = 7.312 \times 1 0^{14} s^{- 1} \to \overset{v}{ˉ} = \frac{v}{c} = 2437200 m^{- 1}$
$\overset{v}{ˉ}$ in $c m^{- 1} = 24372 c m^{- 1}$
For Ist excited state $n = 2$
$24372 = 109677 [\frac{1}{4} - \frac{1}{n _{2}^{2}}]$
$n_{2} = 6$ , i.e., $5^{th}$ excited state