SATHEE: Modern Physics 4 Question 1

Modern Physics 4 Question 1

1. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 \AA$ ). The de-Broglie wavelength of this electron is

(Main 2019, 12 April II)

(a) $3.5 \AA$

(b) $6.6 \AA$

(d) $9.7 \AA$

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Solution:

By Bohr’s IInd postulate, for revolving electron,

$\begin{matrix} Angular momentum = \frac{n h}{2 π} \Rightarrow m v r_{n} = \frac{n h}{2 π} \\ \Rightarrow Momentum of electron, p = m v = \frac{n h}{2 π r_{n}} \end{matrix}$

de-Broglie wavelength associated with electron is

$\begin{aligned} λ_{n} = \frac{h}{p} = \frac{2 π r_{n}}{n} \\ Given, & n & = 3, r_{n} = 4.65 \AA \\ ∴ & λ_{n} & = \frac{(2 \times π \times 4.65)}{3} \approx 9.7 \AA \end{aligned}$

Modern Physics 4 Question 1

1. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 \AA$ ). The de-Broglie wavelength of this electron is

Solution:

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Modern Physics 4 Question 1

1. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65\AA ). The de-Broglie wavelength of this electron is

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 \AA$ ). The de-Broglie wavelength of this electron is