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The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]
- a)2.4 V
- b)– 1.2 V
- c)– 2.4 V
- d)1.2 V
Correct answer is option 'D'. Can you explain this answer?
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The potential difference that must be applied tostop the fastest photo...
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2.01 V
c)1.99 V
d)1.24 V
We can use the formula:
$K_{max} = hf - \phi$
where $K_{max}$ is the maximum kinetic energy of the photoelectron, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal.
First, we need to find the frequency of the incident light:
$c = f\lambda$
where $c$ is the speed of light and $\lambda$ is the wavelength of the light. We are given that the wavelength is 200 nm, so:
$f = \frac{c}{\lambda} = \frac{3\times10^8\text{ m/s}}{200\times10^{-9}\text{ m}} = 1.5\times10^{15}\text{ Hz}$
Now we can calculate the maximum kinetic energy of the photoelectron:
$K_{max} = hf - \phi = (6.63\times10^{-34}\text{ J s})(1.5\times10^{15}\text{ Hz}) - (5.01\text{ eV})(1.6\times10^{-19}\text{ J/eV}) = 1.86\times10^{-19}\text{ J}$
Finally, we can find the potential difference needed to stop this photoelectron:
$V = \frac{K_{max}}{q} = \frac{1.86\times10^{-19}\text{ J}}{1.6\times10^{-19}\text{ C/eV}} = 1.16\text{ V}$
However, this is the potential difference needed to stop an electron at rest. The fastest photoelectron will have a larger kinetic energy, so we need a larger potential difference. We can estimate that the fastest photoelectron will have roughly twice the kinetic energy of the average photoelectron, so we need roughly twice the potential difference. This gives us a final answer of:
$V \approx 2.32\text{ V}$
which is closest to option (a) 2.4 V.
c)1.99 V
d)1.24 V
We can use the formula:
$K_{max} = hf - \phi$
where $K_{max}$ is the maximum kinetic energy of the photoelectron, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal.
First, we need to find the frequency of the incident light:
$c = f\lambda$
where $c$ is the speed of light and $\lambda$ is the wavelength of the light. We are given that the wavelength is 200 nm, so:
$f = \frac{c}{\lambda} = \frac{3\times10^8\text{ m/s}}{200\times10^{-9}\text{ m}} = 1.5\times10^{15}\text{ Hz}$
Now we can calculate the maximum kinetic energy of the photoelectron:
$K_{max} = hf - \phi = (6.63\times10^{-34}\text{ J s})(1.5\times10^{15}\text{ Hz}) - (5.01\text{ eV})(1.6\times10^{-19}\text{ J/eV}) = 1.86\times10^{-19}\text{ J}$
Finally, we can find the potential difference needed to stop this photoelectron:
$V = \frac{K_{max}}{q} = \frac{1.86\times10^{-19}\text{ J}}{1.6\times10^{-19}\text{ C/eV}} = 1.16\text{ V}$
However, this is the potential difference needed to stop an electron at rest. The fastest photoelectron will have a larger kinetic energy, so we need a larger potential difference. We can estimate that the fastest photoelectron will have roughly twice the kinetic energy of the average photoelectron, so we need roughly twice the potential difference. This gives us a final answer of:
$V \approx 2.32\text{ V}$
which is closest to option (a) 2.4 V.
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The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer?
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The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer?.
The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The potential difference that must be applied tostop the fastest photoelectrons emitted by anickel surface, having work function 5.01 eV,when ultraviolet light of 200 nm falls on it, mustbe: [2010]a)2.4 Vb)– 1.2 Vc)– 2.4 Vd)1.2 VCorrect answer is option 'D'. Can you explain this answer?.
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