What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from an energy level with n=4 to an energy level with n=2 ? What is the colour of the radiation ?

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Solution

We have,

Energy level $n = 4$ and $n = 2$

According to the Balmer series

Wave no $= R H [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

Here $n_{1} = 2, n_{2} = 4$

Putting the value then we get,

Wave number $= 109678 (\frac{1}{2^{2}} - \frac{1}{4^{2}})$

$= \frac{109678 \times 3}{16}$

Also,

$λ = \frac{1}{w a v e n o}$

Therefore,

$λ = \frac{16}{109678 \times 3} = 486 n m$

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What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

What is the wavelength of the light emitted when an electron in a hydrogen atom undergoes transition from the energy level n = 4 to energy level n = 2? What is the colour corresponding to their wavelengths? (Given $R_{H} = 109677 c m^{- 1}$ )

Q. 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}

)

Q. Calculate the wavelength of radiation emitted when an electron in a hydrogen atom makes a transition from an energy level with n= 3 to a level with n= 2:

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What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from an energy level with n=4 to an energy level with n=2 ? What is the colour of the radiation ?

We have,Energy level n=4 and n=2According to the Balmer seriesWave no =RH[1n21−1n22]Here n1=2,n2=4Putting the value then we get,Wave number =109678(122−142)=109678×316Also,λ=1wavenoTherefore,λ=16109678×3=486nm