Question

If a photon of energy $7.08\times {10}^{-19}J$ falls on lithium metal having work function $2.42eV$ then the kinetic energy of the ejected election will be:

• A. $1.5eV$
• B. $2.0eV$
• C. $3.0eV$
• D. $2.5eV$

Answer 1

Answer: (B)

$2.0eV$

Photoelectric equation is given by,

$K.E=h\nu -w$

Where $w$=work function, $h\nu =$energy of the photon

$h\nu =7.08\times {10}^{-19}J=4.43eV$, $w=2.42eV$

Now substituting the values ,

$K.E=4.43eV-2.42eV=2.01eV$

$K.E=2eV$