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Question: The angular momentum of an electron in a Bohr orbit of H - atom is \(\text{4}\text{.2178 }\times \te...

The angular momentum of an electron in a Bohr orbit of H - atom is 4.2178 × 1034 kgm2/sec\text{4}\text{.2178 }\times \text{ 1}{{\text{0}}^{-34}}\text{ kg}{{\text{m}}^{2}}/\sec . Calculate the wavelength of the spectral line emitted when an electron falls from this level to the next lower level.

Explanation

Solution

Bohr gave the expression for the angular momentum of an electron i.e.mvr = nh2π\text{mvr = }\dfrac{\text{nh}}{2\pi }. Here, m is the mass of an electron, v is the velocity, r is the radius of the orbit whereas n is the orbit in which the electron is present and h is the Planck's constant.

Complete step by step answer:
-It is given in the question that the angular momentum of an electron is 4.2178 × 1034 kgm2/sec\text{4}\text{.2178 }\times \text{ 1}{{\text{0}}^{-34}}\text{ kg}{{\text{m}}^{2}}/\sec and the value of h is fixed and constant for every element i.e. 6.626 × 1034\text{6}\text{.626 }\times \text{ 1}{{\text{0}}^{-34}} and also the value of pi is 3.14 for every equation.
-Now, we have to find the value of n that is the orbit in which the electron is present.
-Now we know that the angular momentum is the product of the distance of the object from the rotational axis and the linear momentum of the electron.
-So, the formula of angular momentum becomes mvr.

-Now, we know that mvr = nh2π\text{mvr = }\dfrac{\text{nh}}{2\pi } so, the value of n will be:
4.2178 × 1034 = n × 6.626 × 10342 × 3.144.2178\text{ }\times \text{ 1}{{\text{0}}^{-34}}\text{ = n }\times \text{ }\dfrac{6.626\text{ }\times \text{ 1}{{\text{0}}^{-34}}}{2\text{ }\times \text{ 3}\text{.14}}
n = 4\text{n = 4}

-Now, we have to calculate the wavelength of the spectral line when it travels from the 4 energy level to the 3 energy level.
-So, by using the Rydberg formula:
1λ = RH(1n12 - 1n22)\dfrac{1}{\lambda }\text{ = }{{\text{R}}_{\text{H}}}\left( \dfrac{1}{\text{n}_{1}^{2}}\text{ - }\dfrac{1}{\text{n}_{2}^{2}} \right)
1λ = 109678(132 - 142)\dfrac{1}{\lambda }\text{ = 109678}\left( \dfrac{1}{{{3}^{2}}}\text{ - }\frac{1}{{{4}^{2}}} \right)
λ = 1.8 × 104cm\lambda \text{ = 1}\text{.8 }\times \text{ 1}{{\text{0}}^{-4}}\text{cm}
Therefore, the wavelength of the spectral line is 1.8 × 104cm\text{1}\text{.8 }\times \text{ 1}{{\text{0}}^{-4}}\text{cm}.

Note: The Rydberg formula is the transition of the different energies that occur between the energy level. The electron emits a photon when they move from a high energy level to the low energy level. There are many series in the spectral line such as Lyman series, Balmer series, Paschen series, etc