Question

When light of wavelength lesser than 6000 A is
incident on a metal, electrons are emitted. The
approximate work-function of the metal is :
(1) 1 eV
(2) 2eV
(3) 4 eV
(4) 6 eV​

Answer 1

Given:

When light of wavelength lesser than 6000 A° is incident on a metal, electrons are emitted.

To find:

Approximate work function of the metal ?

Calculation:

The work function (threshold energy) is the minimum energy of the photons required to emit an electron from the metal surface.

Let work function be W_(0):

$\therefore \: W_{0} = h \nu$

$\implies\: W_{0} = h \times \dfrac{c}{ \lambda}$

$\implies\: W_{0} = 6.63 \times {10}^{ - 34} \times \dfrac{3 \times {10}^{8} }{ 6000 \times {10}^{ - 10} }$

$\implies\: W_{0} = 6.63 \times {10}^{ - 34} \times \dfrac{3 \times {10}^{8} }{ 6\times {10}^{ - 7} }$

$\implies\: W_{0} = 6.63 \times {10}^{ - 34} \times \dfrac{ {10}^{8} }{ 2\times {10}^{ - 7} }$

$\implies\: W_{0} = 6.63 \times {10}^{ - 34} \times \dfrac{1}{2} \times {10}^{15}$

$\implies\: W_{0} = 3.31 \times {10}^{ - 34} \times {10}^{15}$

$\implies\: W_{0} = 3.31 \times {10}^{ - 19} \: joule$

$\implies\: W_{0} = \dfrac{ 3.31 \times {10}^{ - 19}}{e} \: eV$

$\implies\: W_{0} = \dfrac{ 3.31 \times {10}^{ - 19}}{1.6 \times {10}^{ - 19} } \: eV$

$\implies\: W_{0} = 2 \: eV$

So, work function of the metal is 2 eV.