A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. [2023]

Question

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $x \times 10^{15} Hz$ . The value of $x$ is _______. (Given $h = 4.25 \times 10^{- 15} eV s$ ) [2023]

Smart Achievers · Accepted Answer

(3)

$6 = {}^{4}C_{2} \Rightarrow n_{2} = 4$

$h ν = E_{4} - E_{1}$

$∴$ $ν = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}}) \times \frac{1}{4.25 \times 10^{- 15}} = 3 \times 10^{15} Hz$

Solution

Ans.
(3)

$6 = {}^{4}C_{2} \Rightarrow n_{2} = 4$

$h ν = E_{4} - E_{1}$

$∴$ $ν = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}}) \times \frac{1}{4.25 \times 10^{- 15}} = 3 \times 10^{15} Hz$

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Solution

Q. A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x×1015 Hz. The value of x is _______. (Given h=4.25×10-15 eV s) [2023]

Ans. (3) 6=C24 ⇒ n2=4 hν=E4-E1 ∴ ν=13.6(112-142)×14.25×10-15=3×1015 Hz

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Ans.
(3)

$6 = {}^{4}C_{2} \Rightarrow n_{2} = 4$

$h ν = E_{4} - E_{1}$

$∴$ $ν = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}}) \times \frac{1}{4.25 \times 10^{- 15}} = 3 \times 10^{15} Hz$