Question

Draw a neat and labelled energy level diagram and explain Balmer series and Brackett series of spectral lines for hydrogen atom.
The work function for a metal surface is 2.2 eV. If light of wavelength 5000 $\stackrel{o}{A}$ is incident on the surface of the metal, find the threshold frequency and incident frequency. Will there be an emission of photoelectrons or not?
($c=3\times {10}^{8}m/s$, 1 eV = $1.6\times {10}^{-19}$ J, $h=6.63\times {10}^{-34}J.s.$)

Answer 1

When any pure gas is energised by electrically charging it, by heating or any other method, then its electrons get excited and they jump into orbits of higher energy levels. This is known as excited state.

Later on they come to lower energy level orbits by releasing energy in the form of photons. These photons get released in the form of electromagnetic rays. Due to this we get a spectrum of rays around the gas chamber.

While releasing energy, some of these electrons jump to the intermediate orbits step by step and then come to the orginal orbit, releasing rays in steps. Some electrons jump directly to the original orbit. This state in which electrons are not at all excited is known as ground state.

The wavelength of the emitted rays is given by $\frac{1}{\lambda }=R(\frac{1}{p^2}-\frac{1}{n^2}).$ Here p is the number of lower energy orbit while n is that of the higher. R is Rydberg's constant.

A person named Balmer first tried this experiment with hydrogen gas and observed some visible rays around the hydrogen chamber. These rays were the ones that were jumping to the 2nd level orbit from any higher orbit.

Thus for $p=2$ we get a series of visible rays. $\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2}).$ This visible series is known as Balmer series.

However Balmer failed to notice the invisible rays that were also emitted by the hydrogen chamber. Electrons that jump in the fourth orbit from some higher orbit release rays that lie in the middle infra-red spectrum. For $p=4$ this series is known as Brackett series. $\frac{1}{\lambda }=R(\frac{1}{4^2}-\frac{1}{n^2}).$

Please note that there are other series existing for other values of p.

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work function$={\varphi }_{{}_0}=2.2eV=2.2\times 1.6\times {10}^{-19}Joules$

Incident wavelength$=\lambda =5000A^o=5\times {10}^{-7}metres$

$c=3\times {10}^{8}m/s$ and $h=6.63\times {10}^{-34}Js$

${\upsilon }_{{}_0}=?$ and $\upsilon =?$

$\upsilon =\frac{c}{\lambda }=\frac{3\times {10}^8}{5\times {10}^{-7}}=6\times {10}^{14}Hz$

${\varphi }_{{}_0}=h{\upsilon }_{{}_0}$ $\therefore {\upsilon }_{{}_0}=\frac{{\varphi }_{{}_0}}{h}=\frac{2.2\times 1.6\times {10}^{-19}}{6.63\times {10}^{-34}}=0.531\times {10}^{15}$

$\therefore {\upsilon }_{{}_0}=5.31\times \times {10}^{14}Hz$

Since $\upsilon >{\upsilon }_{{}_0},$ photo-electrons will be emitted.