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Atoms & Nuclei

JEE Main level, numerical answer, medium

56 of 239

Question 56 of 239

For hydrogen atom, ${\lambda }_{1}$ and ${\lambda }_{2}$ are the wavelengths corresponding to the transitions $1$ and $2$ respectively as shown in figure. The ratio of ${\lambda }_{1}$ and ${\lambda }_{2}$ is $\dfrac{x}{32}$ . The value of $x$ is ______.

Show answer and solution

Answer: 27

Wavelength corresponding to the photon emitted in transition

n=3ton=1

\dfrac{1}{{\lambda }_{1}}=R\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{3}^{2}}\right)=\dfrac{8R}{9}

....(1)
Wavelength corresponding to the photon emitted for transition in transition

n=2ton=1

\dfrac{1}{{\lambda }_{2}}=R\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{2}^{2}}\right)=\dfrac{3R}{4}

....(2)

\dfrac{{\lambda }_{1}}{{\lambda }_{2}}=\dfrac{3R}{4}\times \dfrac{9}{8R}=\dfrac{27}{32}

Hence, the value of

x=27

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JEE Main Physics practice question on Atoms & Nuclei (Bohr's Atomic Model), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

For hydrogen atom, λ1{\lambda }_{1}λ1​ and λ2{\lambda }_{2}λ2​ are the wavelengths corresponding to the transitions 111 and 222 respectively as shown in figure. The ratio of λ1{\lambda }_{1}λ1​ and λ2{\lambda }_{2}λ2​ is x32\dfrac{x}{32}32x​. The value of xxx is ______.

JEE Main question paper PDF

More Physics practice

For hydrogen atom, ${\lambda }_{1}$ and ${\lambda }_{2}$ are the wavelengths corresponding to the transitions $1$ and $2$ respectively as shown in figure. The ratio of ${\lambda }_{1}$ and ${\lambda }_{2}$ is $\dfrac{x}{32}$ . The value of $x$ is ______.