Question 356: Dual Nature of Matter and Radiation

Question

Photons of energy $$10 \mathrm{~eV}$$ are incident on a photosensitive surface of threshold frequency $$2 	imes 10^{15} \mathrm{~Hz}$$. The kinetic energy in $$\mathrm{eV}$$ of the photoelectrons emitted is
[Planck's constant $$\mathrm{h}=6.63 	imes 10^{34} \mathrm{~Js}$$ ]

Accepted Answer

First, let's recall that the kinetic energy (KE) of the ejected electrons can be found using the formula:
$$ KE = E - W = h
u - h
u_0 $$
where:
$$ KE $$ is the kinetic energy of the ejected electrons,
$$ E = h
u $$ is the energy of the incident photon,
$$ h $$ is Planck's constant,
$$ 
u $$ is the frequency of the incident photon,
$$ W = h
u_0 $$ is the work function of the material, and
$$ 
u_0 $$ is the threshold frequency.
To find the kinetic energy in electron-volts (eV), we first need to calculate the energy of the incident photons and the work function using the provided values. The energy of the incident photon (E) is given to be $$10 \mathrm{~eV}$$ directly.
The work function (W), however, needs to be calculated. It is the energy corresponding to the threshold frequency ($$
u_0$$), so we can calculate it using:
$$ W = h
u_0 $$
Given that $$h = 6.63 	imes 10^{-34} \mathrm{~Js}$$ (or meter2 kg / s in SI units), and $$
u_0 = 2 	imes 10^{15} \mathrm{~Hz}$$, let's calculate W. However, since our final answer is desired in electron-volts, we must remember that $$1 \mathrm{~eV} = 1.602 	imes 10^{-19} \mathrm{~J}$$.
To calculate the work function in eV, we use:
$$ W = h
u_0 = (6.63 	imes 10^{-34} \mathrm{~Js}) 	imes (2 	imes 10^{15} \mathrm{~Hz}) $$
$$ W = 1.326 	imes 10^{-18} \mathrm{~J} $$
Convert W to eV:
$$ W = \frac{1.326 	imes 10^{-18} \mathrm{~J}}{1.602 	imes 10^{-19} \mathrm{~eV/J}} = 8.29 \mathrm{~eV} $$
Now, we can calculate the kinetic energy of the photoelectrons:
$$ KE = E - W = 10 \mathrm{~eV} - 8.29 \mathrm{~eV} $$
$$ KE = 1.71 \mathrm{~eV} $$
Therefore, the kinetic energy of the photoelectrons emitted is
1.71 eV
, which corresponds to Option D.

Answer

8.29 eV

Answer

6.5 eV

Answer

4.2 eV

Answer

1.71 eV

Dual Nature of Matter and Radiation

Photons of energy $10 \mathrm{~eV} are incident on a photosensitive surface of threshold frequency 2 \times 10^{15} \mathrm{~Hz}. The kinetic energy in \mathrm{eV} of the photoelectrons emitted is [Planck's constant \mathrm{h}=6.63 \times 10^{34} \mathrm{~Js}$ ]

Choose the correct answer: