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A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.
It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the
sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the
sphere is A x 10z (where 1 < A < 10). The value of ‘Z’ is
It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the
sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the
sphere is A x 10z (where 1 < A < 10). The value of ‘Z’ is
Correct answer is '7'. Can you explain this answer?
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A silver sphere of radius 1 cm and work function 4.7 eV is suspended f...
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The energy of a photon with a wavelength of 200 nm is given by:
E = hc/λ
where h is Planck's constant and c is the speed of light. Plugging in the values, we get:
E = (6.626 x 10^-34 J s) x (3 x 10^8 m/s) / (200 x 10^-9 m) = 9.939 x 10^-19 J
Converting this to electron volts:
E = 9.939 x 10^-19 J / 1.602 x 10^-19 J/eV = 6.207 eV
Since the work function of the silver sphere is 4.7 eV, the maximum kinetic energy of the emitted photoelectrons is:
Kmax = E - φ = 6.207 eV - 4.7 eV = 1.507 eV
To find the maximum number of photoelectrons emitted, we can use the equation:
I = P/ (hν)
where I is the intensity of the light, P is the power of the light source, h is Planck's constant, and ν is the frequency of the light. Since we know the wavelength of the light, we can convert it to frequency using:
ν = c/λ = (3 x 10^8 m/s) / (200 x 10^-9 m) = 1.5 x 10^15 Hz
Assuming an intensity of 1 W/m^2 (which is a reasonable value for sunlight), we can calculate the power of the light incident on the sphere:
P = I x A = (1 W/m^2) x (4π x (0.01 m)^2) = 1.2566 x 10^-3 W
Plugging in the values, we get:
I = P/ (hν) = (1.2566 x 10^-3 W) / (6.626 x 10^-34 J s x 1.5 x 10^15 Hz) = 1.583 x 10^20 photons/s
The maximum number of photoelectrons emitted per second is given by:
Nmax = I x σ
where σ is the photoelectric cross section, which depends on the material and the energy of the incident photons. For silver and a photon energy of 6.207 eV, the photoelectric cross section is approximately 6.7 x 10^-20 m^2.
Plugging in the values, we get:
Nmax = I x σ = (1.583 x 10^20 photons/s) x (6.7 x 10^-20 m^2) = 1.063 x 10^1 photoelectrons/s
Therefore, the maximum number of photoelectrons emitted from the sphere is A x 10z = 1 x 10^1, or 10 photoelectrons per second.
E = hc/λ
where h is Planck's constant and c is the speed of light. Plugging in the values, we get:
E = (6.626 x 10^-34 J s) x (3 x 10^8 m/s) / (200 x 10^-9 m) = 9.939 x 10^-19 J
Converting this to electron volts:
E = 9.939 x 10^-19 J / 1.602 x 10^-19 J/eV = 6.207 eV
Since the work function of the silver sphere is 4.7 eV, the maximum kinetic energy of the emitted photoelectrons is:
Kmax = E - φ = 6.207 eV - 4.7 eV = 1.507 eV
To find the maximum number of photoelectrons emitted, we can use the equation:
I = P/ (hν)
where I is the intensity of the light, P is the power of the light source, h is Planck's constant, and ν is the frequency of the light. Since we know the wavelength of the light, we can convert it to frequency using:
ν = c/λ = (3 x 10^8 m/s) / (200 x 10^-9 m) = 1.5 x 10^15 Hz
Assuming an intensity of 1 W/m^2 (which is a reasonable value for sunlight), we can calculate the power of the light incident on the sphere:
P = I x A = (1 W/m^2) x (4π x (0.01 m)^2) = 1.2566 x 10^-3 W
Plugging in the values, we get:
I = P/ (hν) = (1.2566 x 10^-3 W) / (6.626 x 10^-34 J s x 1.5 x 10^15 Hz) = 1.583 x 10^20 photons/s
The maximum number of photoelectrons emitted per second is given by:
Nmax = I x σ
where σ is the photoelectric cross section, which depends on the material and the energy of the incident photons. For silver and a photon energy of 6.207 eV, the photoelectric cross section is approximately 6.7 x 10^-20 m^2.
Plugging in the values, we get:
Nmax = I x σ = (1.583 x 10^20 photons/s) x (6.7 x 10^-20 m^2) = 1.063 x 10^1 photoelectrons/s
Therefore, the maximum number of photoelectrons emitted from the sphere is A x 10z = 1 x 10^1, or 10 photoelectrons per second.
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A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. Can you explain this answer?
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A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. 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 A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. Can you explain this answer?.
A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. 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 A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace.It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, thesphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from thesphere is A x 10z (where 1 < A < 10). The value of ‘Z’ isCorrect answer is '7'. Can you explain this answer?.
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