the energy difference of two states of an atom is 8.1 kev . determine the wavelength of the light emittited or absorbed when an electron jumps Between these two states?
Answers
Explanation:
Correct option is
B
486 nm
According to Balmer formula
Wave number =R
H
[1/n
1
2
–1/n
2
2
]
Here n
1
=2,n
2
=4,R
H
=109678
Putting these values in the equation we get
Wave number =109678(1/2
2
−1/4
2
)=109678×3/16
Also λ= 1/ wave number
Therefore λ=16/109678×3=486nm
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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