Question

The speed of electron revolving in the $4^{th}$ orbit of a hydrogen like atom or ion is $1094km/s$. The atom or ions is:

• A. $H$
• B. $H{e}^{+}$
• C. $L{i}^{2+}$
• D. $B{e}^{3+}$

Answer 1

The correct option is B $H{e}^{+}$

$v_n=v_0\times z/n$

$v_0=2.188\times {10}^6$

1094 km/s = 1094000 m/s

$n=4$

Substituting the above values,

$1094000=2.188\times {10}^6\times z/4$

$z=1094000/547\times {10}^3$

$z=2$

So for $z=2$, it must be $He$ atom.