1) It is observed that when a metal surface is illuminated by light of wavelengt
ID: 1624837 • Letter: 1
Question
1) It is observed that when a metal surface is illuminated by light of wavelength ?, photoelectrons are emitted from the metal at a certain rate. (This rate of electron emission is called the photocurrent.)
(i) If the intensity, but not the wavelength, of the light is increased,
a) there is no change in the photocurrent.
b) the kinetic energy of the photoelectrons increases, but not the photocurrent.
c) there is an increase in both the kinetic energy of the photoelectrons and the photocurrent.
d) the photocurrent increases.
(ii) To increase the kinetic energy of the photoelectrons,
a) use light of a longer wavelength
b) use light of a shorter wavelength.
c) increase the intensity of the light.
2) The figure below shows photoelectric effect data for three different metals. The vertical axis is the maximum kinetic energy of a photoelectron (this is the kinetic energy of the electron when it is just emitted from the cathode), and the horizontal axis is the frequency of the incident light.
(i) The horizontal (frequency)-axis intercepts (when the maximum KE is zero) represent
a) the cutoff (or threshold) wavelengths of the metals
b) the cutoff (or threshold) frequencies of the metals
c) Planck's constant
d) the work functions of the metals
(b) The units for the slopes of these lines is
a) J
b) J·s
c) J/s
(c) The slopes of these lines represent
a) the cutoff (or threshold) frequencies of the metals
b) the work functions of the metals
c) Planck's constant
d) the cutoff (or threshold) wavelengths of the metals
(d) The vertical axis intercepts (where the frequency is zero) represent (in magnitude)
a) Planck's constant
b) the work functions of the metals
c) the cutoff (or threshold) wavelengths of the metals
d) the cutoff (or threshold) frequencies of the metals
Explanation / Answer
(1)
(i) option (d) is correct. The photocurrent increases.
When intensity is increased, the number of photons hitting the metal surface per second will increase. That will increase the number of photoelectrons emitted. so the photocurrent increases.
(ii) option (b) is correct.
The kinetic energy is directly proportional to the frequency. Thus for higher kinetic energy, we need to use higher frequency. Higher frequency means shorter wavelength.
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