Question
Consider $${3^{rd}}$$ orbit of $$H{e^ + }$$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be (given $$K = 9 \times {10^9}$$ constant, $$Z = 2$$ and $$h$$ (Planck's constant) $$ = 6.6 \times {10^{ - 34}}J{\text{ - }}s$$ )
A.
$$2.92 \times {10^6}m/s$$
B.
$$1.46 \times {10^6}m/s$$
C.
$$0.73 \times {10^6}m/s$$
D.
$$3.0 \times {10^8}m/s$$
Answer :
$$1.46 \times {10^6}m/s$$
Solution :
Energy of electron in the 3rd orbit of $$H{e^ + }$$ is
$$\eqalign{
& {E_3} = - 13.6 \times \frac{{{Z^2}}}{{{n^2}}}eV = - 13.6 \times \frac{4}{{{3^2}}}eV \cr
& = - 13.6 \times \frac{4}{9} \times 1.6 \times {10^{ - 19}}\,J \cr} $$
From Bohr’s model,
$$\eqalign{
& {E_3} = - K{E_3} = - \frac{1}{2}{m_e}{v^2} \cr
& \Rightarrow \frac{1}{2} \times 9.1 \times {10^{ - 31}} \times {v^2} = - 13.6 \times \frac{4}{9} \times 1.6 \times {10^{ - 19}} \cr
& \Rightarrow {v^2} = \frac{{136 \times 16 \times 4 \times 2 \times {{10}^{ - 11}}}}{{9 \times 91}} \cr
& {\text{or,}}\,\,v = 1.46 \times {10^6}\;m/s \cr} $$