Question

The photoelectric effect is applied in some burglar alarms. A beam of light shines on a metal electrode in a vacuum tube(photoelectric cell) and causes electrons to be photoelectrically ejected from the surface of the metal. This electrode thus becomes the cathode. The ejected electrons are attracted towards the anode in the vacuum tube and the electrical circuit is completed by means of a battery. If the light beam is blocked by a burglar's arm, the electrical circuit is broken. This sets off the alarm system. What is the maximum wavelength of light that you could use for a burglar alarm if the cathode of the photoelectric cell is made of tungsten and electrons are ejected from tungsten with a kinetic energy of $8.0\times {10}^{-12}$ erg when the wavelength of the incident light is exactly$1.25\times {10}^3\mathring{A}$?

• A. $2.5\times {10}^3\mathring{A}$
• B. $3.4\times {10}^3\mathring{A}$
• C. $4.5\times {10}^3\mathring{A}$
• D. $5.6\times {10}^3\mathring{A}$

Answer 1

Answer: (A)

$2.5\times {10}^3\mathring{A}$

The formula for kinetic energy used is,

$KE=h\nu -h{\nu }_o=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_o}]$

Where ${\lambda }_o$ is the maximum wavelength used.
$h=6.625\times {10}^{-27}erg.s$ and $c=3\times {10}^{10}cm.s^{-1}$

Putting values in the equation,

$8\times {10}^{-12}=\frac{6.625\times {10}^{-27}\times 3\times {10}^{10}}{1.25\times {10}^3\times {10}^{-8}}-\frac{6.625\times {10}^{-27}\times 3\times {10}^{10}}{{\lambda }_o}$

$\Rightarrow {\lambda }_o=2.5\times {10}^{-5}cm=2.5\times {10}^3\times {10}^{-8}cm=2.5\times {10}^3\mathring{A}$