Atomic StructureHard
Question
What should be the increase in kinetic energy of electron in order to decrease its de Broglie wavelength from 100 nm to 50 nm?
Options
A.0.451 keV
B.4.51 × 10−4 eV
C.4.51 × 10−3 eV
D.0.0451 eV
Solution
$\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \Delta E = \frac{h^{2}}{2m}\left( \frac{1}{\lambda_{2}^{2}} - \frac{1}{\lambda_{1}^{2}} \right)$
$= \frac{\left( 6.626 \times 10^{- 34} \right)^{2}}{2 \times 9 \times 10^{- 31}}\left( \frac{1}{\left( 50 \times 10^{- 9} \right)^{2}} - \frac{1}{\left( 100 \times 10^{- 9} \right)^{2}} \right) $$$= 7.24 \times 10^{- 23}\text{ J = 4.5} \times \text{1}\text{0}^{- 4}\text{ eV}$$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The de Broglie wavelength of electron of He+ ion is 3.32 Å. If the photon emitted upon de-excitation of this He+ ion is ...Which of the following will not change if the choice of the zero potential energy is changed in Bohr model -...If an electron in H-atom jumps from one orbit to other, its angular momentum doubles. The distance of electron from nucl...Suppose that a hypothetical atom gives a red, green, blue and violet line spectrum. Which jump according to figure would...Which onr of the following sets of quantum numbers represents an impossible arrangement ?...