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The wavelength or radiations emitted is ...

The wavelength or radiations emitted is `lambda_(0)` when an electron in hydrogen atom jumps from the third orbit to second. If in the H-atom itself, the electron jumps from fourth orbit to second orbit, the wavelength of emitted radiation will be

A

`(20)/(27)lambda_(0)`

B

`(16)/(25)lambda_(0)`

C

`(27)/(20)lambda_(0)`

D

`(29)/(16)lambda_(0)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the Rydberg formula for the hydrogen atom, which relates the wavelengths of emitted radiation to the transitions of electrons between energy levels. The formula is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( \lambda \) is the wavelength of the emitted radiation, - \( R \) is the Rydberg constant, - \( n_1 \) is the principal quantum number of the lower energy level, - \( n_2 \) is the principal quantum number of the higher energy level. ### Step 1: Calculate the wavelength \( \lambda_0 \) for the transition from the third orbit to the second orbit. For the transition from \( n_2 = 3 \) to \( n_1 = 2 \): \[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] Calculating the terms: \[ \frac{1}{\lambda_0} = R \left( \frac{1}{4} - \frac{1}{9} \right) \] Finding a common denominator (36): \[ \frac{1}{\lambda_0} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \] Thus, we have: \[ \lambda_0 = \frac{36}{5R} \] ### Step 2: Calculate the wavelength \( \lambda \) for the transition from the fourth orbit to the second orbit. For the transition from \( n_2 = 4 \) to \( n_1 = 2 \): \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] Calculating the terms: \[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \] Finding a common denominator (16): \[ \frac{1}{\lambda} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \] ### Step 3: Relate \( \lambda \) to \( \lambda_0 \). Now substituting \( R \) from the first equation into the second: From \( \lambda_0 = \frac{36}{5R} \), we can express \( R \) as: \[ R = \frac{36}{5\lambda_0} \] Substituting this into the equation for \( \lambda \): \[ \frac{1}{\lambda} = \frac{36}{5\lambda_0} \cdot \frac{3}{16} \] This simplifies to: \[ \frac{1}{\lambda} = \frac{108}{80\lambda_0} = \frac{27}{20\lambda_0} \] Thus, we find: \[ \lambda = \frac{20\lambda_0}{27} \] ### Final Answer: The wavelength of the emitted radiation when the electron jumps from the fourth orbit to the second orbit is: \[ \lambda = \frac{20\lambda_0}{27} \]
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When electron in hydrogen atom jumps from second orbit to first orbit, the wavelength of radiation emitted is lambda . When electron jumps from third orbit to first orbit, the wavelength of emitted radiation would be _______________ .

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Knowledge Check

  • The wavelength of radiation emitted is lamda_0 when an electron in a hydrogen atom jumps from 3rd to 2nd orbit . If the same hydrogen atom , the electron jumps from 4th orbit to 2nd orbit , then the wavelength of the emitted radiation will be

    A
    `10/25 lamda_0`
    B
    `25/16 lamda_0`
    C
    `27/20 lamda_0`
    D
    `20/27 lamda_0`
  • The wavelength of radiation emitted is lambda_(0) when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of hydrogen atom, the wavelength of radiation emitted will be

    A
    `(16)/(25) lambda_(0)`
    B
    `(20)/(27)lambda_(0)`
    C
    `(27)/(20)lambda_(0)`
    D
    `(25)/(16)lambda_(0)`
  • when an electron jumps from the fourth orbit to the second orbit, one gets the

    A
    second line of paschen series
    B
    second line of balmer serie
    C
    first line pfund series
    D
    second line of lyman series
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