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If the threshold wavelength for a certain metal is 2000 Å, then the work-function of the metal is [1995]
- a)6.2 J
- b)6.2 eV [
- c)6.2 MeV
- d)6.2 keV
Correct answer is option 'B'. Can you explain this answer?
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If the threshold wavelength for a certain metal is 2000 Å, then ...
Threshold wavelength (λ) = 2000 Å = 2000 ×
10–10 m. Work function
10–10 m. Work function
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If the threshold wavelength for a certain metal is 2000 Å, then ...
Angstroms, what is the maximum kinetic energy of the emitted electrons when the metal is illuminated with radiation of wavelength 1500 angstroms?
We can use the following formula to calculate the maximum kinetic energy of the emitted electrons:
K.E. = hν - Φ
where K.E. is the maximum kinetic energy, h is Planck's constant, ν is the frequency of the radiation, and Φ is the work function of the metal (which is related to the threshold wavelength by Φ = hc/λ).
First, we need to find the frequency of the radiation with a wavelength of 1500 angstroms:
ν = c/λ = (3.00 x 10^8 m/s)/(1500 x 10^-10 m) = 2.00 x 10^15 Hz
Next, we can calculate the work function of the metal:
Φ = hc/λ = (6.63 x 10^-34 J s)(3.00 x 10^8 m/s)/(2000 x 10^-10 m) = 9.945 x 10^-19 J
Now we can plug in these values to find the maximum kinetic energy:
K.E. = hν - Φ = (6.63 x 10^-34 J s)(2.00 x 10^15 Hz) - 9.945 x 10^-19 J = 1.32 x 10^-18 J
Note that this answer is in joules, but we can convert it to electron volts (eV) by dividing by the elementary charge (e = 1.60 x 10^-19 C):
K.E. = (1.32 x 10^-18 J)/(1.60 x 10^-19 C) = 8.25 eV
Therefore, the maximum kinetic energy of the emitted electrons is 8.25 electron volts.
We can use the following formula to calculate the maximum kinetic energy of the emitted electrons:
K.E. = hν - Φ
where K.E. is the maximum kinetic energy, h is Planck's constant, ν is the frequency of the radiation, and Φ is the work function of the metal (which is related to the threshold wavelength by Φ = hc/λ).
First, we need to find the frequency of the radiation with a wavelength of 1500 angstroms:
ν = c/λ = (3.00 x 10^8 m/s)/(1500 x 10^-10 m) = 2.00 x 10^15 Hz
Next, we can calculate the work function of the metal:
Φ = hc/λ = (6.63 x 10^-34 J s)(3.00 x 10^8 m/s)/(2000 x 10^-10 m) = 9.945 x 10^-19 J
Now we can plug in these values to find the maximum kinetic energy:
K.E. = hν - Φ = (6.63 x 10^-34 J s)(2.00 x 10^15 Hz) - 9.945 x 10^-19 J = 1.32 x 10^-18 J
Note that this answer is in joules, but we can convert it to electron volts (eV) by dividing by the elementary charge (e = 1.60 x 10^-19 C):
K.E. = (1.32 x 10^-18 J)/(1.60 x 10^-19 C) = 8.25 eV
Therefore, the maximum kinetic energy of the emitted electrons is 8.25 electron volts.
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If the threshold wavelength for a certain metal is 2000 Å, then the work-function of the metal is [1995]a)6.2 Jb)6.2 eV [c)6.2 MeVd)6.2 keVCorrect answer is option 'B'. Can you explain this answer?
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