Q. If a metal has a work function of 2 eV, what is the minimum wavelength of light required to cause the photoelectric effect?
A.620 nm
B.310 nm
C.1240 nm
D.500 nm
Solution
Using the equation λ = hc/E, where E = 2 eV = 2 * 1.6 x 10^-19 J, we find the minimum wavelength λ = (6.63 x 10^-34 J·s * 3 x 10^8 m/s) / (2 * 1.6 x 10^-19 J) = 310 nm.
Q. If a metal has a work function of 2 eV, what is the threshold wavelength for the photoelectric effect?
A.620 nm
B.400 nm
C.500 nm
D.300 nm
Solution
The threshold wavelength can be calculated using the equation λ = hc/W. Substituting h = 4.14 x 10^-15 eV·s, c = 3 x 10^8 m/s, and W = 2 eV gives λ = 620 nm.
Q. If a metal has a work function of 4 eV, what is the minimum wavelength of light required to cause the photoelectric effect?
A.310 nm
B.620 nm
C.1240 nm
D.2480 nm
Solution
The minimum wavelength can be calculated using the equation λ = hc/W. Substituting h = 4.14 x 10^-15 eV·s, c = 3 x 10^8 m/s, and W = 4 eV gives λ = 310 nm.
Q. If the frequency of incident light is doubled, what happens to the kinetic energy of the emitted electrons?
A.It remains the same
B.It doubles
C.It quadruples
D.It decreases
Solution
The kinetic energy of the emitted electrons is given by KE = hf - φ. If the frequency is doubled, the kinetic energy increases as it is directly proportional to frequency.
Q. If the intensity of light is doubled while keeping the frequency constant, what happens to the number of emitted electrons in the photoelectric effect?
A.It doubles
B.It remains the same
C.It is halved
D.It becomes zero
Solution
Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.
Q. If the intensity of light is doubled while keeping the frequency constant, what happens to the number of emitted electrons?
A.It doubles
B.It remains the same
C.It is halved
D.It becomes zero
Solution
Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.
Q. If the work function of a metal is 4.5 eV, what is the threshold wavelength for the photoelectric effect?
A.400 nm
B.500 nm
C.600 nm
D.700 nm
Solution
The threshold wavelength can be calculated using the equation λ = hc/φ. Substituting h = 4.14 x 10^-15 eV·s, c = 3 x 10^8 m/s, and φ = 4.5 eV gives λ ≈ 400 nm.