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On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x
Correct answer is '11'. Can you explain this answer?
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Verified Answer
On a photosensitive metal of area 1 cm2 and work function 2eV, light o...
Energy of photon. E =
= 4eV > 2eV (so photoelectric effect will take place)
= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule
No. of photons falling per second
No. of photoelectron emitted per second
= 4eV > 2eV (so photoelectric effect will take place)= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule
No. of photons falling per second
No. of photoelectron emitted per second
Most Upvoted Answer
On a photosensitive metal of area 1 cm2 and work function 2eV, light o...
To solve this problem, we can use the equation for the number of photoelectrons emitted:
Number of photoelectrons = (Intensity of light) x (Area of metal) x (Time) x (Photoelectric current)
First, we need to convert the intensity of light from watts per square meter to photons per second per square meter. We can use the equation:
Intensity (in photons per second per square meter) = (Intensity in watts per square meter) / (Energy per photon)
The energy per photon can be calculated using the equation:
Energy per photon = Planck's constant x (Speed of light) / (Wavelength)
Given that the intensity of light is 6.4 W/m^2, we need to calculate the energy per photon:
Energy per photon = (6.626 x 10^-34 J s) x (3 x 10^8 m/s) / (400 x 10^-9 m) = 4.97 x 10^-19 J
Now we can calculate the intensity of light in photons per second per square meter:
Intensity (in photons per second per square meter) = (6.4 W/m^2) / (4.97 x 10^-19 J) ≈ 1.29 x 10^19 photons/s/m^2
Next, we can calculate the number of photoelectrons emitted using the equation:
Number of photoelectrons = (Intensity of light) x (Area of metal) x (Time) x (Photoelectric current)
Given that the area of the metal is 1 cm^2 (which is equal to 1 x 10^-4 m^2), and the work function is 2 eV (which is equal to 2 x 1.6 x 10^-19 J), we need to calculate the number of photoelectrons:
Number of photoelectrons = (1.29 x 10^19 photons/s/m^2) x (1 x 10^-4 m^2) x (1 s) x (Photoelectric current)
Since we don't have the value for the photoelectric current, we cannot calculate the number of photoelectrons.
Number of photoelectrons = (Intensity of light) x (Area of metal) x (Time) x (Photoelectric current)
First, we need to convert the intensity of light from watts per square meter to photons per second per square meter. We can use the equation:
Intensity (in photons per second per square meter) = (Intensity in watts per square meter) / (Energy per photon)
The energy per photon can be calculated using the equation:
Energy per photon = Planck's constant x (Speed of light) / (Wavelength)
Given that the intensity of light is 6.4 W/m^2, we need to calculate the energy per photon:
Energy per photon = (6.626 x 10^-34 J s) x (3 x 10^8 m/s) / (400 x 10^-9 m) = 4.97 x 10^-19 J
Now we can calculate the intensity of light in photons per second per square meter:
Intensity (in photons per second per square meter) = (6.4 W/m^2) / (4.97 x 10^-19 J) ≈ 1.29 x 10^19 photons/s/m^2
Next, we can calculate the number of photoelectrons emitted using the equation:
Number of photoelectrons = (Intensity of light) x (Area of metal) x (Time) x (Photoelectric current)
Given that the area of the metal is 1 cm^2 (which is equal to 1 x 10^-4 m^2), and the work function is 2 eV (which is equal to 2 x 1.6 x 10^-19 J), we need to calculate the number of photoelectrons:
Number of photoelectrons = (1.29 x 10^19 photons/s/m^2) x (1 x 10^-4 m^2) x (1 s) x (Photoelectric current)
Since we don't have the value for the photoelectric current, we cannot calculate the number of photoelectrons.
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On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. Can you explain this answer?
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On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. 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 On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. Can you explain this answer?.
On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. 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 On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x Correct answer is '11'. Can you explain this answer?.
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