In a hypothetical Bohr hydrogen atom,the mass | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The radius of the $n$-th orbit in Bohr's model is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$. Since $r \propto 1/m$,if the mass $m$ is

Vedclass · Accepted Answer

$E_0 = -27.2 	ext{ eV}; r_0 = a_0/2$

Vedclass · Answer

(A) The radius of the $n$-th orbit in Bohr's model is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$. Since $r \propto 1/m$,if the mass $m$ is doubled,the new radius $r_0$ becomes $r_0 = a_0/2$.The energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$. Since $E \propto m$,if the mass $m$ is doubled,the new energy $E_0$ becomes $E_0 = 2 	imes (-13.6 	ext{ eV}) = -27.2 	ext{ eV}$.Thus,the correct option is $A$.

In a hypothetical Bohr hydrogen atom,the mass of the electron is doubled. The energy $E_0$ and the radius $r_0$ of the first orbit will be ($a_0$ is the Bohr radius).

Similar Questions

The time of revolution of an electron around a nucleus of charge $Ze$ in the $n^{th}$ Bohr orbit is directly proportional to

Bohr’s second postulate implies the quantisation of:

Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason : According to classical physics all moving electrons radiate.

Assertion $(A)$: The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principal quantum number $(n)$.
Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to $n^{x}$,where $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module