A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n = 4$ level. Determine the wavelength and frequency of photon.
For ground level, $n_{1}=1$
Let $E_{1}$ be the energy of this level. It is known that $E_{1}$ is related with $n_{1}$ as
$E_{1}=\frac{-13.6}{n_{1}^{2}}\, e V$
$=\frac{-13.6}{1^{2}}=-13.6\, e V$
The atom is excited to a higher level, $n_{2}=4$
Let $E_{2}$ be the energy of this level.
$\therefore E_{2}=\frac{-13.6}{n_{2}^{2}} \,e V$
$=\frac{-13.6}{4^{2}}=-\frac{13.6}{16}\, e V$
The amount of energy absorbed by the photon is given as
$E=E_{2}-E_{1}$
$=\frac{-13.6}{16}-\left(-\frac{13.6}{1}\right)$
$=\frac{13.6 \times 15}{16} \,e V$
$=\frac{13.6 \times 15}{16} \times 1.6 \times 10^{-19}=2.04 \times 10^{-18} \,J$
For a photon of wavelength $\lambda,$ the expression of energy is written as
$E=\frac{h c}{\lambda}$
Where, $h=$ Planck's constant $=6.6 \times 10^{-34} \,Js$
$c=$ Speed of light $=3 \times 10^{8} \,m / s$
$\therefore \lambda=\frac{h c}{E}$
$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}}$
$=9.7 \times 10^{-8}\, m =97\, nm$
And, frequency of a photon is given by the relation, $v=\frac{c}{\lambda}$
$=\frac{3 \times 10^{8}}{9.7 \times 10^{-8}} \approx 3.1 \times 10^{15} \,Hz$
Hence, the wavelength of the photon is $97\, nm$ while the frequency is $3.1 \times 10^{15} \,Hz$.
Write equation of mass of photon.
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.
Assertion : The photoelectrons produced by a monochromatic light beam incident on a metal surface, have a spread in their kinetic energies.
Reason : The work function of the metal varies as a function of depth from the surface.
Using the Heisenberg uncertainty principle, arrange the following particles in the order of increasing lowest energy possible.
$(I)$ An electron in $H _{2}$ molecule
$(II)$ A hydrogen atom in a $H _{2}$ molecule
$(III)$ A proton in the carbon nucleus
$(IV)$ $A H _{2}$ molecule within a nanotube
According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is