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प्रश्न
How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?
उत्तर
Balmer series contains wavelengths ranging from 364 nm (for n2 = 3) to 655 nm (n_2 = ∞).
So, the given range of wavelength (380−780 nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by
`1/lamda = R (1/2^2 - 1/n^2)`
Here, R = Rydberg's constant = `1.097 xx 10^7 m^-1`
The wavelength for the transition from n = 3 to n = 2 is given by
`1/lamda_1 = R(1/2^2 - 1/3^2)`
`lamda_1 = 656.3 nm`
The wavelength for the transition from n = 4 to n = 2 is given by
`1/lamda_2 = R(1/2^2 - 1/4^2)`
`lamda_2 =486.1 nm`
The wavelength for the transition from n = 5 to n = 2 is given by
`1/lamda_3 = R (1/2^2 - 1/5^2)`
`lamda_3 = 434.0 nm`
The wavelength for the transition from n = 6 to n = 2 is given by
`1/lamda_4 = R (1/2^2 - 1/6^2)`
`lamda_4 = 410.2 nm`
The wavelength for the transition from n = 7 to n = 2 is given by
`1/lamda_5 = R (1/2^2 - 1/7^2)`
`lamda_5 = 397.0 nm`
Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.
Lyman series contains wavelengths ranging from 91 nm (for n2 = 2) to 121 nm (n_2 = ∞)
So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series.
The wavelength in the Lyman series can be found by
`1/lamda = R(1/1^2 - 1/2^2)`
The wavelength for the transition from n = 2 to n = 1 is given by
`1/lamda_1 = R(1/1^2 - 1/2^2)`
`lamda_1 = 122 nm`
The wavelength for the transition from n = 3 to n = 1 is given by
`1/lamda^2 = R(1/1^2 - 1/2^2)`
`lamda_2 = 103 nm`
The wavelength for the transition from n = 4 to n = 1 is given by
`1/lamda^3= R(1/1^2 - 1/4^2)`
`lamda_3 = 97.3 nm`
The wavelength for the transition from n = 5 to n = 1 is given by
`1/lamda_4 = R(1/1^2 - 1/5^2)`
`lamda_4 = 95.0 nm`
The wavelength for the transition from n = 6 to n = 1 is given by
`1/lamda_5 = R(1/1^2 - 1/6^2)`
`lamda_5 = 93.8 nm`
So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.
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