Modern Physics Ques 66
- Ultraviolet light of wavelengths $800 $ $\AA$ and $700$ $ \AA$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy $1.8 $ $eV$ and $4.0$ $ eV$ respectively. Find the value of Planck’s constant.
$(1983,4 M)$
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Answer:
Correct Answer: 66.$(6.6 \times 10^{-34} J-s)$
Solution:
- When $800 $ $\AA$ wavelength falls on hydrogen atom (in ground state) $13.6 $ $eV$ energy is used in liberating the electron. The rest goes to kinetic energy of electron.
Hence, $K=E-13.6$ (in $eV$ ) or
$\left(1.8 \times 1.6 \times 10^{-19}\right)=\frac{h c}{800 \times 10^{-10}}-13.6 \times 1.6 \times 10^{-19}$ $\quad$ …….(i)
Similarly, for the second wavelength :
$\left(4.0 \times 1.6 \times 10^{-19}\right)=\frac{h c}{700 \times 10^{-10}}-13.6 \times 1.6 \times 10^{-19}$ $\quad$ …….(ii)
Solving these two equations, we get
$ h \approx 6.6 \times 10^{-34} J-s $
BETA
