Question

The energy difference of two states of an atom is 8.1 keV. Determine the wavelength of
he light emitted or absorbed when an electron jumps between these two states.​

Answer 1

Explanation:

As we know,

1/λ=R(1/n

1

2

−1/n

2

2

)

=R(1/3

2

−1/4

2

)

λ=1.875 ×10

−6

m

ΔE=hc/λ = 1.06×10

−19

Answer 2

Answer:

The wavelength of the light emitted or absorbed when an electron jumps between these two states is 0.153nm

Explanation:

Given,

The energy difference between two energy levels of an atom (E) = 8.1 kev

To find,

The wavelength of the light emitted or absorbed when an electron jumps Between these two states (λ)

Concept,

The energy of an electronic transition is given by:

E = hc/λ

Where, h = plack's constant = 6.64 × 10⁻³⁴ J.s

c = speed of light = 3 × 10⁸ m/s

λ = Wavelength of the light emitted or absorbed

Calculation,

When an electron makes a transition from a higher energy level to a lower energy level then the electron releases some energy equal to the energy difference between the energy levels.

Similarly, if an electron jumps from a lower energy level to a higher energy level then the electron absorbs the energy equal to the difference in the energy levels.

So, E = hc/λ

⇒ 8.1 kev = 6.64 × 10⁻³⁴ × 3 × 10⁸/λ

⇒ 8.1 × 10³ × 1.6 × 10⁻¹⁹ = 6.64 × 10⁻³⁴ × 3 × 10⁸/λ (As 1 ev =  1.6 × 10⁻¹⁹J)

⇒ λ = 0.153 × 10⁻⁹ m

⇒ λ = 0.153 nm

Therefore, the wavelength of the light emitted or absorbed when an electron jumps between these two states is 0.153nm.

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