Question 140: Dual Nature of Matter and Radiation

Question

The electric field at a point associated with a light wave is given by
E = 200 [sin (6 $$	imes$$ 1015)t + sin (9 $$	imes$$ 1015)t] Vm$$-$$1
Given : h = 4.14 $$	imes$$ 10$$-$$15 eVs
If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

Accepted Answer

Frequency of EM waves = $${6 \over {2\pi }} 	imes {10^{15}}$$ and $${9 \over {2\pi }} 	imes {10^{15}}$$
Energy of one photon of these waves
$$ = \left( {4.14 	imes {{10}^{ - 15}} 	imes {6 \over {2\pi }} 	imes {{10}^{15}}} ight)$$ eV
and $$\left( {4.14 	imes {{10}^{ - 15}} 	imes {9 \over {2\pi }} 	imes {{10}^{15}}} ight)$$ eV
= 3.95 eV and 5.93 eV
$$\Rightarrow$$ Energy of maximum energetic electrons
= 5.93 $$-$$ 2.50 = 3.43 eV

Answer

1.90 eV

Answer

3.27 eV

Answer

3.60 eV

Answer

3.42 eV

Dual Nature of Matter and Radiation

The electric field at a point associated with a light wave is given by E = 200 [sin (6 $\times 1015)t + sin (9 \times 1015)t] Vm-1 Given : h = 4.14 \times 10-$15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

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