Question

The threshold frequency for metal is \[{10^{15}}Hz\]. If the light of wavelength \[{\rm{4000}}\mathop {\rm{A}}\limits^{\rm{0}} \] falls on its surface, then which of the following statement is correct?
A. Photoelectrons come out with a velocity of 100 m/sec
B. No photoelectric emission takes place
C. Photoelectrons come out with a velocity of 124 m/sec
D. Photoelectrons come out with a velocity of 9.8 m/sec

Answer 1

Hint:When light is incident on a metal surface, electrons are emitted known as photoelectrons. This phenomenon is called the photoelectric effect. The minimum frequency under which the emission of photoelectrons takes place is known as the threshold frequency.

Formula Used:
To find the frequency of light of wavelength is,
\[\nu = \dfrac{c}{\lambda }\]
Where, c is the speed of light and \[\lambda \] is wavelength of light.

Complete step by step solution:
The frequency of light of wavelength is,
\[\nu = \dfrac{c}{\lambda }\]
Substitute the value of c as \[3 \times {10^8}m{s^{ - 1}}\] and \[\lambda {\rm{ = 4000}}\mathop {\rm{A}}\limits^{\rm{0}} \] then we get,
\[\nu = \dfrac{{3 \times {{10}^8}}}{{4000 \times {{10}^{ - 10}}}}\]
\[\therefore \nu = 0.75 \times {10^{15}}Hz\]
They have given the value of threshold frequency, that is \[{v_0} = {10^{15}}Hz\]. There is a condition for which the photoelectric effect will take place, that is the incident frequency should be greater than the threshold frequency, \[v > {v_0}\] then the photoelectric emission will take place. But here if we compare these two frequencies, \[v < {v_0}\]. Therefore, no photoelectric emission takes place.

Hence, Option B is the correct answer.

Note:There is an alternate method to find the solution that is given below.
The formula to find the threshold wavelength is,
\[{\lambda _0} = \dfrac{c}{{{f_0}}}\]
Here, \[{f_0}\] is threshold frequency, that is \[{10^{15}}Hz\] and c is speed of light, \[3 \times {10^8}\]
Now, substitute the value in above equation we get,
\[{\lambda _0} = \dfrac{{3 \times {{10}^8}}}{{{{10}^{15}}}}\]
\[{\lambda _0} = 3 \times {10^{ - 7}}m\]
\[\therefore {\lambda _0} = 3000\mathop A\limits^0 \]
Since, frequency and wavelength are inversely proportional, \[{\lambda _0} > \lambda \] then the photoelectric emission will take place.
Here, \[\lambda > {\lambda _0}\]
Therefore, no photoelectric emission takes place.