If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, the emitted radiation has a frequency (R = Rydberg constant., C = velocity of light)

$\frac{8 R C}{9}$

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$\frac{3 R C}{27}$

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$\frac{5 R C}{36}$

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$\frac{R C}{25}$

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Solution

The correct option is C
$\frac{5 R C}{36}$

$v = \frac{c}{λ} = R C [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = R C [\frac{1}{4} - \frac{1}{9}] = \frac{5 R C}{36}$

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Similar questions

Q. If an electron in hydrogen atom jumps from an orbit of level

n = 3

to an orbit of level

n = 2

, the emitted radiation has a frequency

(R =

Rydberg constant,

C =

velocity of light)

Q. If an electron in hydrogen atom jumps from an orbit of level

n = 3

to an orbit of level

n = 2

, emitted radiation has a frequency.
(R = Rydberg's constant, c = velocity of light)

If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, the emitted radiation has a frequency (R = Rydberg constant., C = velocity of light)

If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, then emitted radiation has a frequency equal to,

Q. If the electron in a hydrogen atom jumps from an orbit with level

n_{1} = 2

to an orbit with level

n_{2} = 1

the emitted radiation has a wavelength given by:

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If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, the emitted radiation has a frequency (R = Rydberg constant., C = velocity of light)

The correct option is C 5RC36 v=cλ=RC[1n21−1n22]=RC[14−19]=5RC36

The correct option is C
$\frac{5 R C}{36}$

$v = \frac{c}{λ} = R C [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = R C [\frac{1}{4} - \frac{1}{9}] = \frac{5 R C}{36}$