What is the threshold frequency for the photoelectric effect on lithium (phi= 2.93eV)? What is the stopping potential if the wavelength of the incident light is 400nm?
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To determine the threshold frequency for the photoelectric effect on lithium, you need to compute the energy of the threshold frequency, which can be done using the equation:
E = h * f
where E is the energy, h is Planck's constant (6.626 x 10^-34 J*s), and f is the frequency.
The threshold frequency corresponds to the minimum frequency of light required for the photoelectric effect to occur. In this case, since we are given the work function (phi) instead of the threshold frequency directly, we can use the equation:
phi = h * f_threshold
Rearranging the equation, we find:
f_threshold = phi / h
Substituting the given values:
phi = 2.93 eV = 2.93 * 1.6 x 10^-19 J (since 1 eV = 1.6 x 10^-19 J)
Now we can calculate the threshold frequency:
f_threshold = (2.93 * 1.6 x 10^-19 J) / (6.626 x 10^-34 J*s)
f_threshold ≈ 7.10 x 10^14 Hz
To determine the stopping potential (V_stop) when the wavelength of the incident light is 400 nm, you can use the equation:
V_stop = (hc / λ) - (phi / e)
where V_stop is the stopping potential, λ is the wavelength of light, h is Planck's constant, c is the speed of light, phi is the work function, and e is the elementary charge.
Substituting the given values:
λ = 400 nm = 400 x 10^-9 m
phi = 2.93 eV = 2.93 * 1.6 x 10^-19 J
c = 3 x 10^8 m/s
Now we can calculate the stopping potential:
V_stop = ((6.626 x 10^-34 J*s) * (3 x 10^8 m/s)) / (400 x 10^-9 m) - ((2.93 * 1.6 x 10^-19 J) / (1.6 x 10^-19 C))
V_stop ≈ 0.84 V
Therefore, the threshold frequency for the photoelectric effect on lithium is approximately 7.10 x 10^14 Hz, and the stopping potential when the wavelength of the incident light is 400 nm is approximately 0.84 V.