An electron in a Bohr orbit of hydrogen atom with the quantum number $$n2$$ has an angular momentum $$4.2176$$×$$10$$−$$34kgm2s$$−$$1$$. If the electron drops from this level to the next lower level, the wavelength of this line is

Correct option is 41.876 x 104 Å

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An electron in a Bohr orbit of hydrogen atom with the quantum number n₂ has an angular momentum $4.2176 \times 10^{- 34} {kgm}^{2} s^{- 1}$ . If the electron drops from this level to the next lower level, the wavelength of this line is

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18 nm

187 .6 pm

1876 Å

1.876 x 104 Å

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detailed solution

Correct option is D

Angular momentum, L=4.2176×10−34=n2h2π⇒ n2=4For the transition from n2 = 4 to n1 = 3, the wavelength of spectral line λ.1λ=13.6hc132−142=13.6eV1240eVnm79×16λ=1240×14413.6×7=1876nm=18760Å =1.876×104Å

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An electron in a Bohr orbit of hydrogen atom with the quantum number n2 has an angular momentum 4.2176×10−34kgm2s−1. If the electron drops from this level to the next lower level, the wavelength of this line is

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free study material

An electron in a Bohr orbit of hydrogen atom with the quantum number n₂ has an angular momentum $4.2176 \times 10^{- 34} {kgm}^{2} s^{- 1}$ . If the electron drops from this level to the next lower level, the wavelength of this line is