The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be

(Given ionization energy of $H = 2.18 \times 10^{- 18} J a t o m^{- 1}$ and $h = 6.625 \times 10^{- 34} J s$ )

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Solution

The correct option is A

$E_{i o n i s a t i o n} = E_{\infty} - E_{n} = \frac{13.6 Z_{e f f}^{2}}{n^{2}} e v$

= $[\frac{13.6 Z^{2}}{n_{2}^{2}} - \frac{13.6 Z^{2}}{n_{1}^{2}}]$

$E = h v = \frac{13.6 \times 1^{2}}{(1)^{2}} - \frac{13.6 \times 1^{2}}{(4)^{2}}$ ,

$h v = 13.6 - 0.85$

$∴ h = 6.625 \times 10^{- 34}$

$v = \frac{13.6 - 0.85}{6.625 \times 10^{- 34}} \times 1.6 \times 10^{- 19} = 3.08 \times 10^{15} s^{- 1}$

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Similar questions

The frequency of radiations emitted when electron falls from n = 4, to n = 1, in H-atom would be:

(Given E₁ for H=2.18×10^-18J atom ^-1 and h=6.625 ×10^-34 Js.)

n = 4

n = 1

H = 2.18 \times 10^{- 18} J a t o m^{- 1}

h = 6.625 \times 10^{- 34} J s)

= 1.097 \times 10^{- 2}

^{- 1}

The frequency of radiations emitted when electron falls from n = 4, to n = 1, in H-atom would be:

(Given E₁ for H=2.18×10^-18J atom ^-1 and h=6.625 ×10^-34 Js.)

n = 4

n =

H = 2.18 10^{- 18} J a t o m^{- 1}

h = 6.625 \times 10^{- 34} J s)

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