Question

Calculate frequency and wave length of photon during the transition of electron in hydrogen from n = 6 to n = 2 orbit​

Answer 1

Answer:

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Explanation:

since n1=5andn1=2, this transition gives rise to a spectral line in the visible region of the balmer series. From equation

ΔE=2.18×10−18J[152−122]

4.58×10−19J

it is an emission energy

the frequency of the photon ( taking energy in terms of magnitude ) is given by

v=ΔEh

4.58×10−19J6.626×10−34Js

6.91×1014Hz

λ=cv=3.0×108ms−16.91×1014Hz=434nm

Answer 2

Given: n₁ = 6, n₂ = 2

To find: frequency and wavelength

Solution:

(a) 1/λ = Rh ( 1/n₁² - 1/n₂²)

where λ is wavelength and Rh is the Rydberg constant

1/λ = Rh(1/2² - 1/6²)

1/λ = Rh(2/24)

Rydberg constant value is 10973731 m⁻¹

Therefore λ = 24 / 2 ×10973731 m

                λ = 12/ 10973731 m

                   = 11.009 × 10⁻7 m

                   = 1100.9 nm

Therefore, wavelength is 1100.9 nm

(b) frrequency = speed / wavelength

                        = 3 × 10⁸ / 110.9 × 10⁻⁹

                        = 0.27 × 10 ¹⁷ Hz

Therefore, frequency is 27 × 10¹⁵ Hz