Question

The energy of electron which is moving in the orbit that has its radius sixteen times the radius of first Bohr orbit for H-atom is:

$[Given:E_{1,H}=2.176\times {10}^{-18}J/atoms]$

• A. $-1.36\times {10}^{-19}$ Joules
• B. $-1.36\times {10}^{-20}$ Joules
• C. $-1.36\times {10}^{-22}$Joules
• D. $-1.36\times {10}^{-26}$Joules

Answer 1

Answer: (A)

$-1.36\times {10}^{-19}$ Joules

Energy E = KE + PE

$E=\frac{1}{2}m{v}^2-\frac{kZ{e}^2}{r}$

where $k=\frac{1}{4\pi {ϵ}_o}$

$E=-\frac{kZ{e}^2}{2r}$

So, $E\propto \frac{1}{r}$

Radius is sixteen times the radius of first Bohr orbit for H-atom, energy will decrease by sixteen times.

$[Given:E_{1,H}=2.176\times {10}^{-18}J/atoms]$

$E=-\frac{E_{1,H}}{16}=-\frac{2.176\times {10}^{-18}}{16}=-1.36\times {10}^{-19}J$

Hence, the correct option is $A$