Question

For hydrogen atom, energy of an electron in first excited state is $-3.4 \, \text{eV}$, K.E. of the same electron of hydrogen atom is $x \, \text{eV}$.Value of $x$ is ____ $\times 10^{-1} \, \text{eV}$. (Nearest integer)

Answer 1

The energy of an electron in the first excited state for a hydrogen atom is given by:

$E = -3.4 \, \text{eV}$

For hydrogen, the kinetic energy (K.E.) in an orbit is given by:

$\text{K.E.} = -\frac{E}{2}$

Thus,

$x = -\left(-\frac{3.4}{2}\right) \times 10 = 34 \, \text{eV}$