
The threshold frequency for a certain photosensitive metal is ${v_0}$. When it is illuminated by light of frequency v = 2${v_0}$, the maximum velocity of photoelectrons is ${v_0}$. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency v =5 ${v_0}$?
A). $\sqrt 2 {v_0}$
B). 2${v_0}$
C). 2$\sqrt 2 {v_0}$
D). 4${v_0}$
Answer
538.8k+ views
Hint: The kinetic energy of the electron is proportional to the frequency of the incident radiation about a threshold frequency.
The kinetic energy is independent of the intensity of the radiation.
Complete step by step solution:
The relation between the Kinetic Energy and the frequency with this expression.
$\dfrac{1}{2}m{\left( {{v_{\max }}} \right)^2} = h\left( {v - {v_0}} \right)$ ---- (1)
Here, v is the frequency,
${v_0}$ Is threshold frequency
m is the mass
h is the Planck’s constant
$v _{\text{max}}$ Is the maximum velocity
Given, condition is when it is illuminated by light then the frequency is 2${v_0}$
Now, we put the value of v is equal to 2${v_0}$in equation (1)
$\dfrac{1}{2}m{\left( {{v_0}} \right)^2} = h(2{v_0} - {v_0})$
$\dfrac{1}{2}m{\left( {{v_0}} \right)^2} = h{v_0}$ --- (2)
Now, the new condition is when it is illuminated by light then the frequency is 5${v_0}$
Now, we put the value of v is equal to 5${v_0}$ in equation (1)
$\dfrac{1}{2}m{\left( v \right)^2} = h(5{v_0} - {v_0})$
$\dfrac{1}{2}m{\left( v \right)^2} = 4h{v_0}$ --- (3)
Now, we have value for different condition
Now, we divide (3) by (2)
$\begin{align}
& \dfrac{{\left( {\dfrac{1}{2}m{v^2}} \right)}}{{\left( {\dfrac{1}{2}m{{\left( {{v_0}} \right)}^2}} \right)}} = \dfrac{{4h{v_0}}}{{h{v_0}}} \\
& \Rightarrow \dfrac{{{v^2}}}{{{{\left( {{v_0}} \right)}^2}}} = 4 \\
& \Rightarrow {v^2} = 4{\left( {{v_0}} \right)^2} \\
& \Rightarrow v = \sqrt {4{{\left( {{v_0}} \right)}^2}} \\
& \Rightarrow v = 2{v_0} \\
\end{align} $
So, the relation between the v and ${v_0}$ is v = 2${v_0}$
So, the option (B) is correct
Note: The term threshold frequency is defined as the minimum frequency of the light which causes the electrons to be released from the metal surface when the light falls on it.
The threshold frequency is not depending on the intensity of the light. The intensity of the light affects the photoelectric current.
The frequency increases the kinetic energy of the photoelectron.
The kinetic energy is independent of the intensity of the radiation.
Complete step by step solution:
The relation between the Kinetic Energy and the frequency with this expression.
$\dfrac{1}{2}m{\left( {{v_{\max }}} \right)^2} = h\left( {v - {v_0}} \right)$ ---- (1)
Here, v is the frequency,
${v_0}$ Is threshold frequency
m is the mass
h is the Planck’s constant
$v _{\text{max}}$ Is the maximum velocity
Given, condition is when it is illuminated by light then the frequency is 2${v_0}$
Now, we put the value of v is equal to 2${v_0}$in equation (1)
$\dfrac{1}{2}m{\left( {{v_0}} \right)^2} = h(2{v_0} - {v_0})$
$\dfrac{1}{2}m{\left( {{v_0}} \right)^2} = h{v_0}$ --- (2)
Now, the new condition is when it is illuminated by light then the frequency is 5${v_0}$
Now, we put the value of v is equal to 5${v_0}$ in equation (1)
$\dfrac{1}{2}m{\left( v \right)^2} = h(5{v_0} - {v_0})$
$\dfrac{1}{2}m{\left( v \right)^2} = 4h{v_0}$ --- (3)
Now, we have value for different condition
Now, we divide (3) by (2)
$\begin{align}
& \dfrac{{\left( {\dfrac{1}{2}m{v^2}} \right)}}{{\left( {\dfrac{1}{2}m{{\left( {{v_0}} \right)}^2}} \right)}} = \dfrac{{4h{v_0}}}{{h{v_0}}} \\
& \Rightarrow \dfrac{{{v^2}}}{{{{\left( {{v_0}} \right)}^2}}} = 4 \\
& \Rightarrow {v^2} = 4{\left( {{v_0}} \right)^2} \\
& \Rightarrow v = \sqrt {4{{\left( {{v_0}} \right)}^2}} \\
& \Rightarrow v = 2{v_0} \\
\end{align} $
So, the relation between the v and ${v_0}$ is v = 2${v_0}$
So, the option (B) is correct
Note: The term threshold frequency is defined as the minimum frequency of the light which causes the electrons to be released from the metal surface when the light falls on it.
The threshold frequency is not depending on the intensity of the light. The intensity of the light affects the photoelectric current.
The frequency increases the kinetic energy of the photoelectron.
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