
If the speed of photoelectrons is $ {10^4}m{s^{ - 1}} $ , what should be the frequency of the incident radiation on a potassium metal? Given, work function of potassium $ = 2.3eV $ .
Answer
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Hint :Work function is defined as the minimum energy required to remove an electron from a point on the surface to infinity or to the vacuum. It can also be called a thermodynamic work. The minimum frequency required to start the photoelectric effect is known as threshold frequency.
Complete Step By Step Answer:
We know that $ E $ $ = {W_o} + {E_K} $ , where $ {W_o} $ is the work function and $ {E_K} $ is the kinetic energy of the photoelectron that has been liberated.
As the value of the work function has already been given in the question.
Therefore, $ \Rightarrow $ $ {W_o} = 2.3eV $
To convert it into joules we have to multiply it with the value of the charge of an electron.
So, $ {W_o} = 2.3 \times 1.6 \times {10^{ - 19}} $
$ \Rightarrow $ $ {W_o} = 3.68 \times {10^{ - 19}}J $
And, $ \Rightarrow $ $ {E_K} = \dfrac{1}{2}m{v^2} $
So, $ {E_K} = \dfrac{1}{2} \times 9 \times {10^{ - 31}} \times {\left( {{{10}^4}} \right)^2}J $
$ {E_K} = 4.5 \times {10^{ - 23}}J $
Now putting the values of work function and kinetic energy in the above equation, we get,
$ E = \left( {3.68 \times {{10}^{ - 19}} + 4.5 \times {{10}^{ - 23}}} \right)J $
$ \Rightarrow $ $ E = 3.68045 \times {10^{ - 19}}J $
We know the formula of energy that is $ E = hf $ where f is the frequency of the metal and h is the Planck’s constant.
So, $ f = \dfrac{E}{h} $
$ \Rightarrow $ $ f = \dfrac{{3.68045 \times {{10}^{ - 19}}}}{{6.63 \times {{10}^{ - 34}}}} $
Therefore,
$ \Rightarrow $ $ f = 0.56 \times {10^{15}}Hz $ .
Note :
When an electrically charged particle gets released from a material or within a material after absorbing electromagnetic radiation then this phenomenon is known as photoelectric effect. The other definition of this phenomenon is that when an electron gets ejected from the metal plate when the light falls on it. The energy radiated can be in any form such as it can be radiated in the form of infrared, ultraviolet light, visible light, X-rays and gamma rays.
Complete Step By Step Answer:
We know that $ E $ $ = {W_o} + {E_K} $ , where $ {W_o} $ is the work function and $ {E_K} $ is the kinetic energy of the photoelectron that has been liberated.
As the value of the work function has already been given in the question.
Therefore, $ \Rightarrow $ $ {W_o} = 2.3eV $
To convert it into joules we have to multiply it with the value of the charge of an electron.
So, $ {W_o} = 2.3 \times 1.6 \times {10^{ - 19}} $
$ \Rightarrow $ $ {W_o} = 3.68 \times {10^{ - 19}}J $
And, $ \Rightarrow $ $ {E_K} = \dfrac{1}{2}m{v^2} $
So, $ {E_K} = \dfrac{1}{2} \times 9 \times {10^{ - 31}} \times {\left( {{{10}^4}} \right)^2}J $
$ {E_K} = 4.5 \times {10^{ - 23}}J $
Now putting the values of work function and kinetic energy in the above equation, we get,
$ E = \left( {3.68 \times {{10}^{ - 19}} + 4.5 \times {{10}^{ - 23}}} \right)J $
$ \Rightarrow $ $ E = 3.68045 \times {10^{ - 19}}J $
We know the formula of energy that is $ E = hf $ where f is the frequency of the metal and h is the Planck’s constant.
So, $ f = \dfrac{E}{h} $
$ \Rightarrow $ $ f = \dfrac{{3.68045 \times {{10}^{ - 19}}}}{{6.63 \times {{10}^{ - 34}}}} $
Therefore,
$ \Rightarrow $ $ f = 0.56 \times {10^{15}}Hz $ .
Note :
When an electrically charged particle gets released from a material or within a material after absorbing electromagnetic radiation then this phenomenon is known as photoelectric effect. The other definition of this phenomenon is that when an electron gets ejected from the metal plate when the light falls on it. The energy radiated can be in any form such as it can be radiated in the form of infrared, ultraviolet light, visible light, X-rays and gamma rays.
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