Question

Threshold wavelength for a metal having work function ${\omega _0}$ is $\lambda $ . Then the threshold wavelength for the metal having work function $2{\omega _0}$ is:
A. $4\lambda $
B. $2\lambda $
C. $\dfrac{\lambda }{2}$
D. $\dfrac{\lambda }{4}$

Answer 1

Hint-The work function of a metal is the minimum amount of energy required to make an electron free from the metal surface. If the incoming radiation has an energy equal to the work function of the metal then the frequency of the radiation is known as threshold frequency. The relationship between frequency and wavelength is given as $\upsilon = \dfrac{c}{\lambda }$ , where c is the speed of light and $\lambda $ is the wavelength. By using this we can find the threshold wavelength when the work function is made twice the initial value.

Complete step by step solution:
It is given that the threshold wavelength for a metal having work function ${\omega _0}$ is $\lambda $ .
We need to find the threshold wavelength for the metal having work function $2{\omega _0}$.
The work function is the energy required to break an electron free from the metal surface.
 We know that the threshold frequency is the minimum frequency of the incident photon required for removing an electron from the metal surface. That is the energy of this photon should be equal to the work function of the metal.
 We know that energy of a photon is given as
$E = h\upsilon $................(1)
 Where, E is the energy, h is the Planck’s constant and $\upsilon $ is the frequency.
We know that the speed of light is given as
$c = \upsilon \lambda $
$ \Rightarrow \upsilon = \dfrac{c}{\lambda }$
Where, c is the speed of light and $\lambda $ is the wavelength of the light.
The wavelength corresponding to the threshold frequency is known as threshold wavelength.
On substituting this in equation 1 we get
$ \Rightarrow E = h\dfrac{c}{\lambda }$
At threshold frequency we have
$E = {\omega _0}$
$ \Rightarrow h\dfrac{c}{\lambda } = {\omega _0}$................(2)
Now let us write the equation when the work function is made twice the initial value.
$ \Rightarrow \omega ' = 2{\omega _0}$
 Let the wavelength needed in this case be $\lambda '$
Then, we get
$ \Rightarrow h\dfrac{c}{{\lambda '}} = \omega '$
$ \Rightarrow h\dfrac{c}{{\lambda '}} = 2{\omega _0}$................(3)
 By dividing equation 2 by 3, then we get
$ \Rightarrow \dfrac{{h\dfrac{c}{\lambda }}}{{h\dfrac{c}{{\lambda '}}}} = \dfrac{{{\omega _0}}}{{2{\omega _0}}}$
$\therefore \lambda ' = \dfrac{\lambda }{2}$
 This is the value of threshold wavelength when the work function of the metal $2{\omega _0}$.

So, the correct answer is option C.

Note:Remember that the work function of a metal is the energy required to make an electron free from the surface. Electrons will be emitted only if the incoming radiation has an energy equal to the work function of the metal. If the incident photon is having energy greater than the work function then that extra energy will be converted as the kinetic energy of the emitted electrons.