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The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 X 107 m-1)
- a)91 nm
- b)192 nm
- c)406 nm
- d)9,1 x 10-8 nm
Correct answer is option 'A'. Can you explain this answer?
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The wavelength of the radiation emitted when an electron in a hydrogen atom falls from infinity to stationary state 1 can be determined using the Rydberg formula and the energy levels of the hydrogen atom.
The Rydberg formula is given by:
1/λ = R * (1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of the emitted radiation
- R is the Rydberg constant (1.097 x 10^7 m^-1)
- n₁ is the initial energy level (infinity in this case)
- n₂ is the final energy level (1 in this case)
We need to find the wavelength, so let's substitute the given values into the formula:
1/λ = (1.097 x 10^7 m^-1) * (1/∞² - 1/1²)
Since 1/∞ is defined as 0, the formula simplifies to:
1/λ = (1.097 x 10^7 m^-1) * (0 - 1/1²)
1/λ = (1.097 x 10^7 m^-1) * (0 - 1)
1/λ = -1.097 x 10^7 m^-1
To find the wavelength, we need to take the reciprocal of both sides:
λ = -1 / (-1.097 x 10^7 m^-1)
λ = 1 / (1.097 x 10^7 m^-1)
λ ≈ 91 nm
Therefore, the wavelength of the radiation emitted when the electron falls from infinity to stationary state 1 is approximately 91 nm. Hence, option A is the correct answer.
The Rydberg formula is given by:
1/λ = R * (1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of the emitted radiation
- R is the Rydberg constant (1.097 x 10^7 m^-1)
- n₁ is the initial energy level (infinity in this case)
- n₂ is the final energy level (1 in this case)
We need to find the wavelength, so let's substitute the given values into the formula:
1/λ = (1.097 x 10^7 m^-1) * (1/∞² - 1/1²)
Since 1/∞ is defined as 0, the formula simplifies to:
1/λ = (1.097 x 10^7 m^-1) * (0 - 1/1²)
1/λ = (1.097 x 10^7 m^-1) * (0 - 1)
1/λ = -1.097 x 10^7 m^-1
To find the wavelength, we need to take the reciprocal of both sides:
λ = -1 / (-1.097 x 10^7 m^-1)
λ = 1 / (1.097 x 10^7 m^-1)
λ ≈ 91 nm
Therefore, the wavelength of the radiation emitted when the electron falls from infinity to stationary state 1 is approximately 91 nm. Hence, option A is the correct answer.
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The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 X 107 m-1)a)91 nmb)192 nmc)406 nmd)9,1 x 10-8 nmCorrect answer is option 'A'. Can you explain this answer?
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