Question

Electrons are revolving around the nucleus in $n_1{}^{th}$ orbit of an atom, have atomic number $Z_1$ and in the $n_1{}^{th}$ orbit of other atom, have atomic number $Z_2$ then which os gthe following is correct:
[Where P = Linear momentum, L = Angular momentum, f = frequency fo revolution and K.E. = kinetic energy]

• A. $\frac{L_1}{L_2}=\frac{n_1}{n_2}$
  
• B. $\frac{P_1}{P_2}=\frac{Z_1{n}_2}{Z_2{n}_1}$
  
• C. $\frac{f_1}{f_2}={\left(\frac{Z_2}{Z_1}\right)}^2{\left(\frac{n_1}{n_2}\right)}^3$
  
• D. $\frac{(K.E{)}_1}{(K.E.{)}_2}={\left(\frac{Z_1}{Z_2}\frac{n_2}{n_1}\right)}^2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\frac{L_1}{L_2}=\frac{n_1}{n_2}$

B $\frac{P_1}{P_2}=\frac{Z_1{n}_2}{Z_2{n}_1}$

D $\frac{(K.E{)}_1}{(K.E.{)}_2}={\left(\frac{Z_1}{Z_2}\frac{n_2}{n_1}\right)}^2$

$\frac{L_1}{L_2}=\frac{m{v}_1{r}_1}{m{v}_2{r}_2}=\frac{\frac{Z_1}{n_1}\times \frac{n_1^2}{Z_1}}{\frac{Z_2}{n_2}\times \frac{n_2^2}{Z_2}}=\frac{n_1}{n_2}\Rightarrow \frac{P_1}{P_2}=\frac{m{v}_1}{m{v}_2}=\frac{\frac{Z_1}{n_1}}{\frac{Z_2}{n_2}}=\frac{Z_1{n}_2}{Z_2{n}_1}$

$\frac{f_1}{f_2}=\frac{\frac{v_1}{2\pi r_1}}{\frac{v_2}{2\pi r_2}}=\frac{\frac{Z_1}{n_1}\times \frac{Z_1}{n_1^2}}{\frac{Z_2}{n_2}\times \frac{Z_2}{n_2^2}}={\left(\frac{Z_1}{Z_2}\right)}^2\cdot {\left(\frac{n_2}{n_1}\right)}^3\Rightarrow \frac{K.E{.}_1}{K.E{.}_2}=\frac{\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{v}_2^2}={\left(\frac{Z_1}{Z_2}\right)}^2\times {\left(\frac{n_2}{n_1}\right)}^2={\left(\frac{Z_1{n}_2}{Z_2{n}_1}\right)}^2$