According to Bohr's atomic theory :-A. Kineti | vedclass

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

Question

(D) According to Bohr's theory :$A$. $KE = 13.6 \frac{Z^{2}}{n^{2}} \ eV/atom \Rightarrow KE \propto \frac{Z^{2}}{n^{2}}$ (Correct)$B$. Speed of $e^{-

Vedclass · Accepted Answer

$A$ and $D$ only

Vedclass · Answer

(D) According to Bohr's theory :$A$. $KE = 13.6 \frac{Z^{2}}{n^{2}} \ eV/atom \Rightarrow KE \propto \frac{Z^{2}}{n^{2}}$ (Correct)$B$. Speed of $e^{-} \propto \frac{Z}{n} \Rightarrow v 	imes n \propto Z$ (Incorrect)$C$. Frequency of revolution of $e^{-} = \frac{v}{2 \pi r}$. Since $v \propto \frac{Z}{n}$ and $r \propto \frac{n^{2}}{Z}$,then Frequency $\propto \frac{Z/n}{n^{2}/Z} = \frac{Z^{2}}{n^{3}}$ (Incorrect)$D$. $F = \frac{kZe^{2}}{r^{2}}$. Since $r \propto \frac{n^{2}}{Z}$,then $F \propto \frac{Z}{(n^{2}/Z)^{2}} = \frac{Z^{3}}{n^{4}}$ (Correct)Thus,statements $A$ and $D$ are correct.

Similar Questions

The wavelength of the radiation emitted,when in a hydrogen atom an electron falls from infinity to stationary state $n=1$,would be ............... $nm$ (Rydberg constant $= 1.097 \times 10^7 \ m^{-1}$)

The energy of an electron in the $3^{rd}$ orbit of $H$-atom (in $J$) is approximately:

Ratio of velocities of electrons of hydrogen atom in $1^{st}$,$2^{nd}$,$3^{rd}$ orbit is

Find the wavelength in the Lyman series for $n_2 = 3$ and $n_1 = 1$.

Bohr's model can explain:

Vedclass Test Series

Exam Paper Generator

Online Exam Module