JEE Exam > JEE Questions > Light rays of wavelengths 6000 and of photon...
Start Learning for Free
Light rays of wavelengths 6000 Å and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x 10-34 J − s ; Velocity of light = 3 x 108 ms-1 ]
- a)2 x 1018
- b)10 x 1018
- c)12 x 1017
- d)12 x 1015
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 ...
S, speed of light = 3 x 10^8 m/s, charge of electron = 1.6 x 10^-19 C]
We can start by calculating the energy of a single photon of wavelength 6000 Å using the formula:
E = hc/λ
where h is Planck's constant (6.64 x 10^-34 J s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength in meters.
Converting the wavelength from Å to meters, we get:
λ = 6000 x 10^-10 m = 6 x 10^-7 m
Plugging in the values, we get:
E = (6.64 x 10^-34 J s) x (3 x 10^8 m/s) / (6 x 10^-7 m)
E = 3.32 x 10^-19 J
This is the energy of one photon of wavelength 6000 Å.
Next, we can calculate the number of photons incident per second per unit area using the intensity given:
I = P/A
where I is the intensity (39.6 W/m^2), P is the power (in watts) of the light source, and A is the area (in square meters) of the surface being illuminated.
Since we're only interested in the number of photons incident, we can use the formula:
P = nE/t
where n is the number of photons, E is the energy of one photon (3.32 x 10^-19 J), and t is the time (1 second).
Solving for n, we get:
n = P/t / E
n = (39.6 W/m^2) / (3.32 x 10^-19 J) = 1.19 x 10^20 photons/m^2/s
This is the number of photons incident per second per unit area.
Since only 1% of these photons emit photoelectrons, we can multiply this number by 0.01 to get the number of photoelectrons emitted per second per unit area:
n_e = n x 0.01
n_e = (1.19 x 10^20 photons/m^2/s) x 0.01 = 1.19 x 10^18 electrons/m^2/s
This is the final answer.
We can start by calculating the energy of a single photon of wavelength 6000 Å using the formula:
E = hc/λ
where h is Planck's constant (6.64 x 10^-34 J s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength in meters.
Converting the wavelength from Å to meters, we get:
λ = 6000 x 10^-10 m = 6 x 10^-7 m
Plugging in the values, we get:
E = (6.64 x 10^-34 J s) x (3 x 10^8 m/s) / (6 x 10^-7 m)
E = 3.32 x 10^-19 J
This is the energy of one photon of wavelength 6000 Å.
Next, we can calculate the number of photons incident per second per unit area using the intensity given:
I = P/A
where I is the intensity (39.6 W/m^2), P is the power (in watts) of the light source, and A is the area (in square meters) of the surface being illuminated.
Since we're only interested in the number of photons incident, we can use the formula:
P = nE/t
where n is the number of photons, E is the energy of one photon (3.32 x 10^-19 J), and t is the time (1 second).
Solving for n, we get:
n = P/t / E
n = (39.6 W/m^2) / (3.32 x 10^-19 J) = 1.19 x 10^20 photons/m^2/s
This is the number of photons incident per second per unit area.
Since only 1% of these photons emit photoelectrons, we can multiply this number by 0.01 to get the number of photoelectrons emitted per second per unit area:
n_e = n x 0.01
n_e = (1.19 x 10^20 photons/m^2/s) x 0.01 = 1.19 x 10^18 electrons/m^2/s
This is the final answer.
Free Test
FREE
| Start Free Test |
Community Answer
Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 ...
Given,
Wavelength = 6000Å
Intensity 'I' = 39.6 W/m^2
length 'l' = 1m^2 (per Unit Area)
time 't' = 1s
no. of photons emitted in One second 'n' = ?
Solution :-
no. of photons emitted = Energy Crossing 1m^2 in 1s
Divided by
Energy of each photon
Energy of each photon(w.k.t) = hc/lambda(wavelength)
= (6.64 x 10^-34 x 3 x 10^8)/(6000 x 10^-10)
= 332 x 10^-21 Js
Energy crossing/passing 1m^2 in 1s =
(39.6 W/m^2)x(1m^2)x(1s)
= 39.6Js
no. of photons = ( 39.6Js)/(332 x 10^-21 Js)
= 119.2 x 10^18
(we can simplify)
= 120.2 x 10^18
= 120.2 x 10^18 x 10^-1
= 12 x 10^17 (Answer)
Hope, you got it..
Wavelength = 6000Å
Intensity 'I' = 39.6 W/m^2
length 'l' = 1m^2 (per Unit Area)
time 't' = 1s
no. of photons emitted in One second 'n' = ?
Solution :-
no. of photons emitted = Energy Crossing 1m^2 in 1s
Divided by
Energy of each photon
Energy of each photon(w.k.t) = hc/lambda(wavelength)
= (6.64 x 10^-34 x 3 x 10^8)/(6000 x 10^-10)
= 332 x 10^-21 Js
Energy crossing/passing 1m^2 in 1s =
(39.6 W/m^2)x(1m^2)x(1s)
= 39.6Js
no. of photons = ( 39.6Js)/(332 x 10^-21 Js)
= 119.2 x 10^18
(we can simplify)
= 120.2 x 10^18
= 120.2 x 10^18 x 10^-1
= 12 x 10^17 (Answer)
Hope, you got it..
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer?
Question Description
Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer?.
Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer?.
Solutions for Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Light rays of wavelengths 6000 and of photon intensity 39.6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 ms-1 ]a)2 x1018b)10 x1018c)12 x1017d)12 x1015Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup