What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from the energy level with $n = 4$ to energy level $n = 2$ ?
(given $R_{n}$ = $109678 c m^{- 1}$ )

786 nm

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486 nm

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586 nm

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986 nm

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Solution

The correct option is B 486 nm
According to Balmer formula

Wave number $= R_{H} [1 / n_{1}^{2} - 1 / n_{2}^{2}]$

Here $n_{1} = 2, n_{2} = 4, R_{H} = 109678$

Putting these values in the equation we get

Wave number $= 109678 (1 / 2^{2} - 1 / 4^{2}) = 109678 \times 3 / 16$

Also $λ =$ 1/ wave number

Therefore $λ = 16 / 109678 \times 3 = 486 n m$

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What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from the energy level with n=4 to energy level n=2 ?(given Rn = 109678cm−1 )

The correct option is B 486 nmAccording to Balmer formulaWave number =RH[1/n21–1/n22]Here n1=2,n2=4,RH=109678Putting these values in the equation we getWave number =109678(1/22−1/42)=109678×3/16Also λ= 1/ wave numberTherefore λ=16/109678×3=486nm

What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from the energy level with $n = 4$ to energy level $n = 2$ ?
(given $R_{n}$ = $109678 c m^{- 1}$ )