TS EAMCET 19-July-2022 Shift 2 Solved Paper

Radius of Bohr's orbit in hydrogen atom can be calculated as $r = \frac{n^2 h^2}{4 \pi^2 m e^2} \times \frac{1}{Z}$ or $r=0.529 \times \frac{n^2}{Z}\ \text{\AA}$ where, $n=$ number of the orbit $Z=$ atomic number For hydrogen's third Bohr's orbit, $r_3 = 0.529 \because \frac{(3)^2}{1}\ \text{\AA}$$= 9 \times 0.529\ \text{\AA}$......(i) Also $r_2 = 0.529 \times (2)^2$ (Given, radius of second Bohr's orbit radius.)$r_2 = 4 \times 0.529$ $\frac{r_2}{4}=0.529$.....(ii) From Eqs. (i) and (ii), $r_3 = 9 \times \frac{r_2}{4} \Rightarrow r_3 = \frac{9}{4} r_2$ Hence, radius of third Bohr's orbit $= \frac{9}{4} r_2$Energy of $n$th Bohr's orbit is given as $E_n = \frac{k Z^2}{n^2}$ where, $n=$ number of orbit, $Z=$ atomic number [For hydrogen, $Z=1$ ] $\therefore E_2 = k \frac{(1)^2}{(2)^2}$ [Given, second Bohr's orbit energy.] $k = E_2 \times 4$ $E_2 = \frac{k}{4}$.......(iii) Also, $E_3 = k \frac{(1)^2}{(3)^2} = \frac{k}{9}$.....(iv) From Eqs. (iii) and (iv), $E_3 = \frac{E_2 \times 4}{9} \Rightarrow E_3 = \frac{4}{9} E_2$ Hence, energy of third Bohr's orbit is $\frac{4}{9} E_2$.