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Question: Photoelectric effect is observed from the surface for frequency’s \(3\times {{10}^{14}}Hz\) and \(2\...

Photoelectric effect is observed from the surface for frequency’s 3×1014Hz3\times {{10}^{14}}Hz and 2×1014Hz2\times {{10}^{14}}Hz for the incidence radiation. If the maximum kinetic energies are in ratio 2:1 then the threshold frequency is
A. 1014Hz{{10}^{14}}Hz
B. 32×1014Hz\dfrac{3}{2}\times {{10}^{14}}Hz
C. 43×1014Hz\dfrac{4}{3}\times {{10}^{14}}Hz
D. None of the above

Explanation

Solution

In the question, we are already given the ratio between the maximum kinetic energies. We can solve this question by first applying the formula for the maximum kinetic energies for each frequency and then dividing the two. After dividing, we will get a relation on equating that divided equation with the given ratio.

Complete answer:
Before solving this question, let us take a look at all the given values
The given frequencies are
v1=3×1014Hz{{v}_{1}}=3\times {{10}^{14}}Hz
v2=2×1014Hz{{v}_{2}}=2\times {{10}^{14}}Hz
And,
maximum kinetic energies are in ratio 2:1
so,
Maximum Kinetic Energy, K.E. max ​= hv−hvThv-h{{v}_{T}}
So,
KEmax⁡1K{{E}_{\max 1}} = hv1−hvTh{{v}_{1}}-h{{v}_{T}} …………….. (1)
Now, again
KEmax⁡2K{{E}_{\max 2}}= ​hv2−hvTh{{v}_{2}}-h{{v}_{T}} ……………… (2)
Now, on dividing equation 1 from equation 2
We have,
KEmax⁡1KEmax⁡2=hv1−hvThv2−hvT\dfrac{K{{E}_{\max 1}}}{K{{E}_{\max 2}}}=\dfrac{h{{v}_{1}}-h{{v}_{T}}}{h{{v}_{2}}-h{{v}_{T}}}
So,
⇒21=hv1−hvThv2−hvT\Rightarrow \dfrac{2}{1}=\dfrac{h{{v}_{1}}-h{{v}_{T}}}{h{{v}_{2}}-h{{v}_{T}}}
⇒2hv2−2hvT=hv1−hvT\Rightarrow 2h{{v}_{2}}-2h{{v}_{T}}=h{{v}_{1}}-h{{v}_{T}}
⇒hvT=2hv2−hv1\Rightarrow h{{v}_{T}}=2h{{v}_{2}}-h{{v}_{1}}
⇒vT=2v2−v1\Rightarrow {{v}_{T}}=2{{v}_{2}}-{{v}_{1}}
Taking the given values,
We have
⇒vT=2×2×1014−3×1014\Rightarrow {{v}_{T}}=2\times 2\times {{10}^{14}}-3\times {{10}^{14}}
⇒vT=1014Hz\Rightarrow {{v}_{T}}={{10}^{14}}Hz
So,
If, photoelectric effect is observed from the surface for frequency’s 3×1014Hz3\times {{10}^{14}}Hz and 2×1014Hz2\times {{10}^{14}}Hzfor the incidence radiation. If the maximum kinetic energies are in ratio 2:1 then the threshold frequency is 1014Hz{{10}^{14}}Hz

Therefore, the correct answer to the given question will be Option - A, i.e., 1014Hz{{10}^{14}}Hz .

Note:
As we can see, we have found a general formula for the threshold energy when the similar data is given, i.e.,
⇒vT=v2−v1k−1\Rightarrow {{v}_{T}}=\dfrac{{{v}_{2}}-{{v}_{1}}}{k-1}
Where k is the given ration between the maximum kinetic energies
The frequency of the threshold is defined as the minimum frequency of incident radiation below which photoelectric emissions are not entirely feasible.