Question

An electron jumps from n shell to l shell in the he+ ion calculate the wavelength of radiation emitted

Answer 1

Answer:

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Answer 2

The wavelength of the radiation emitted is 3r/4.

Given:-

Atom = He+

Atomic Number (z) = 2

Lower Shell (n1) = l = 2

Higher Shell (n2) = n = 4

To Find:-

The wavelength of the radiation emitted.

Solution:-

To find out the wavelength of the radiation emitted, you should follow these simple steps.

Now,

Atom = He+

Atomic Number (z) = 2

Lower Shell (n1) = l = 2

Higher Shell (n2) = n = 4

Wavelength (lambda) = ?

r = Rydberg's Constant

$lambda = r \times {z}^{2} (\frac{1}{ {(n1)}^{2} } - \frac{1}{ {(n2)}^{2} } )$

$\gamma = r \times {2}^{2} (\frac{1}{ {2}^{2} } - \frac{1}{ {4}^{2} } )$

$\gamma = 4r \times ( \frac{1}{4} - \frac{1}{16} )$

$\gamma = 4r \times (\frac{1 \times 4 \: - 1}{16} )$

$\gamma = 4r \times \frac{3}{16}$

$\gamma = \frac{3r}{4}$

Hence, The wavelength of the radiation emitted is 3r/4.

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