Question

The radiation is emitted when a hydrogen atom goes from a higher energy state to eluga energy state. The wavelength of one line in visible region of atomic spectrum of hydrogen is $6.63\ast {10}^{-}7m.$ Energy difference between the two state is

• A. $3.0\ast {10}^{-19}J$
• B. $1.0\ast {10}^{-18}J$
• C. $5.0\ast {10}^{-10}J$
• D. $6.5\ast {10}^{-7}J$

Answer 1

The correct option is A $3.0\ast {10}^{-19}J$

$\lambda =6.63\times {10}^{-7}m$

$\Delta E=h\nu =\frac{hc}{\lambda }$

$\Rightarrow \Delta E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{6.63\times {10}^{-7}}$

$\Rightarrow \Delta E=3.0\times {10}^{-19}J$