An electron in an excited state of Li^(2+) ion has angular momentum 3h/(2π) . The de Broglie wavelength of the electron in this state is pπα_0 (where…

Question

An electron in an excited state of Li^(2+) ion has angular momentum 3h/(2π) . The de Broglie wavelength of the electron in this state is pπα_0 (where a_0 is the Bohr radius). The value of p is

MARKS Content Team · Accepted Answer

Correct answer: 2. From Bohr's law mvr=nh/(2π)=3h/(2π) (from eqes.) ⇒n=3 And momentum =mv=3h/(2πr) Now, radius of n^th shell, r=(n^2)/za_0 ⇒r=(3^2)/3.a_0 ∵Z_Li=3 ⇒r=3a_0 From De Broglie law wavelength=h/Momentum ⇒λ=h/mv=h/(3h2π)r ⇒λ=(2πr)/3=(2π)/3×3a_0 λ=2πa_0=pπa_0 ⇒P=2

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 $p π α_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is

Solution

More Dual Nature of Matter Questions

An electron in an excited state of Li2+ ion has angular momentum 3h2π . The de Broglie wavelength of the electron in this state is pπα0 (where a0 is the Bohr radius). The value of p is

Solution

More Dual Nature of Matter Questions

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 $p π α_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is