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A photon of wavelength 5000 Å strikes a metal surface having work function of 2.20 eV. The kinetic energy of the emitted photo electron is
- a)4.4 x 10–20 J
- b)0.425 eV
- c)2.20 eV
- d)No electron will be emitted
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A photon of wavelength 5000 Å strikes a metal surface having wor...
Angstroms has an energy of approximately 2.48 x 10^-19 joules.
This can be calculated using the equation:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.
Converting the wavelength from Angstroms to meters:
5000 Angstroms = 5000 x 10^-10 meters
Plugging in the values:
E = (6.626 x 10^-34 J s) x (2.998 x 10^8 m/s) / (5000 x 10^-10 m)
E = 2.48 x 10^-19 joules
Therefore, the energy of a photon with a wavelength of 5000 Angstroms is approximately 2.48 x 10^-19 joules.
This can be calculated using the equation:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.
Converting the wavelength from Angstroms to meters:
5000 Angstroms = 5000 x 10^-10 meters
Plugging in the values:
E = (6.626 x 10^-34 J s) x (2.998 x 10^8 m/s) / (5000 x 10^-10 m)
E = 2.48 x 10^-19 joules
Therefore, the energy of a photon with a wavelength of 5000 Angstroms is approximately 2.48 x 10^-19 joules.
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Community Answer
A photon of wavelength 5000 Å strikes a metal surface having wor...
Maximum KE=hf - hfo=[ 12400/lamda(inA*) - 2.20 ]ev
= 12400/5000 - 2.20 ev=2.48 -2.20ev
=0.28ev=0.28×1.6×10^-19J
=4.48×10^-20J
= 12400/5000 - 2.20 ev=2.48 -2.20ev
=0.28ev=0.28×1.6×10^-19J
=4.48×10^-20J
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A photon of wavelength 5000 Å strikes a metal surface having work function of 2.20 eV. The kinetic energy of the emitted photo electron isa)4.4 x 10–20 Jb)0.425 eVc)2.20 eVd)No electron will be emittedCorrect answer is option 'A'. Can you explain this answer?
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