SATHEE: Modern Physics 4 Question 24

Modern Physics 4 Question 24

25. An electron in an excited state of $L i^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . The de Broglie wavelength of the electron in this state is $p π a_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is

(2015 Adv.)

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Solution:

Angular momentum $= n \frac{h}{2 π} = 3 \frac{h}{2 π}$

$∴ n = 3$

Now, $r_{n} \propto \frac{n^{2}}{z}$

$∴ r_{3} = \frac{(3)^{2}}{3} (a_{0}) = 3 a_{0}$

Now, $m v_{3} r_{3} = 3 \frac{h}{2 π}$

$∴ m v_{3} (3 a_{0}) = 3 \frac{h}{2 π}$

or $\frac{h}{m v_{3}} = 2 π a_{0}$ or $\frac{h}{P_{3}} = 2 π a_{0} (∵ P = m v)$

or $λ_{3} = 2 π a_{0}$

$λ = \frac{h}{P}$

$∴$ Answer is 2 .

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Modern Physics 4 Question 24

25. An electron in an excited state of Li2+ ion has angular momentum 3h2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is

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25. An electron in an excited state of $L i^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . The de Broglie wavelength of the electron in this state is $p π a_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is