Question

Calculate the energy and frequency of the radiation emitted when an electron jumps from n=3 to n=2 in a hydrogen atom. Plzzz help me​

Answer 1

Answer:

Energy = 6.55 × 10-²⁰ Joules

Frequency = 4.572 × 10¹⁴Hz

Explanation:

∆E = Rh × [ 1/ni² - 1/nf² ]. (In an hydrogen atom)

Given, ni = 3

nf = 2

Since, ni > nf, it is an emission spectrum, or losing of energy.

∆E = 2.18 × 10-¹⁸ [ 1/3² - 1/2² ]

= 2.18 × 10-¹⁸ × (4 - 9) / 36

= 2.18 × 10-¹⁸ × -5/36

= - 0.3027 × 10-¹⁸ Joules

=> -3.03× 10-¹⁹Joules

E = hυ

3.03 × 10-¹⁹ = 6.626 × 10-³⁴ /υ

=> 3.03 × 10-¹⁹/ 6.626 × 10-³⁴ = υ

which is approximately 4.572 × 10¹⁴Hz

Answer 2

Answer:

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