Question

If the speed of electron in first Bohr orbit of hydrogen be ${}^{\prime }{x}^{\prime }$, then speed of the electron in second orbit of $H{e}^{+}$ is :

• A. $x/2$
• B. $2x$
• C. $x$
• D. $4x$

Answer 1

The correct option is C $x$

The speed of an electron in $n^{th}$ Bohr orbit of a H - like atom is given by
$v_n=2.189\times {10}^6\frac{Z}{n}m/s$
$\frac{v_2\left(H{e}^{+}\right)}{v_1\left(H\right)}=\frac{2\times 1}{2\times 1}=1$
$\therefore v_2\left(H{e}^{+}\right)=x$