Question

Explanation As the threshold energy remains constant, change in energy is equal to change in kinetic energy.
Change in kinetic energy is given by ( Delta E=(1 / 2) m v^{wedge} 2=h c / lambda )
( Delta E=left(6.6 times 10^{wedge}-34 times 3 times 10^{wedge} 8right) /left(4.86 times 10^{wedge}-7right) )
( Delta E=4.074 times 10^{wedge}-19 mathrm{J} )
( Delta E=2.546 mathrm{eV} )
Hence, change in kinetic energy and change in total energy both is ( 2.546 mathrm{eV} )

# How much will the kinetic energy and total energy of an electron in H atom change if the atom emits a photon of wavelength 4860Å?

Solution