Question

A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential difference of 1.00 \mathrm{~V} is required to turn back all the ejected electrons, what is the wavelength of the light beam? (a) 355 nm(b) 497 nm(c) 744 nm(d) 1.42 pm(e) none of those answers

Answer 1

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

• Work function is a material property defined as the minimum amount of energy  required to infinitely remove electrons from the surface of a particular solid.
• The potential difference required to support all emitted electrons is called the stopping potential which is given by $v_0=\frac{K.E_m_a_x}{e}$ .....(1)
• where $v_0$ is the stopping potential and e is the charge of the electron given by $1.6\times10^-^1^9$ .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation  is given by:

$K.E_m_a_x=E-\phi$      ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

  $ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV$

As we know that $E=\frac{hc}{\lambda}$  ....(3)

where Speed of light,$c = 3\times10^8 m/s$ and Planck's constant , $h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs$

From equation (3) , we get

$\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm$

Learn about more einstein photoelectric equation  here:

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