A caterpillar climbs up a one-meter wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the caterpil
2 answers:
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Answer:
The number of electrons emitted from the metal per second increases.
Explanation:
The photoelectric effect can be explained by thinking the incident light as made of many photons, each carrying an energy of
where h is the Planck constant and f is the light frequency. A photoelectron is extracted from the metal if the energy of the incoming photon, which hits the electron and gives all its energy, is at least equal to the work function of the metal, . Moreover, each photon of the incident light hits only one electron in the metal.
Given these premises, we can analyze each statement:
The work function of the metal decreases. --> FALSE. The work function is just the energy needed to extract photoelectrons from the metal: so, it depends only on the properties of the metal, and not on the intensity of the incident light.
The number of electrons emitted from the metal per second increases. --> TRUE. The intensity of the incident light is proportional to the number of photons contained in the light: so, the higher the intensity, the larger the number of photons that hit the electrons in the metal, the larger the number of electrons emitted.
The maximum speed of the emitted electrons increases. --> FALSE. The energy of the electrons emitted depends ONLY on the energy of the incoming photon, so it depends only on the frequency of the photon, not on the number of photon (intensity). In fact, only 1 photon at time hits 1 electron, so the intensity of the light does not affect the energy of the electrons (and so, it does not affect their speed)
The stopping potential increases. --> FALSE. The stopping potential is the potential needed to stop the electrons in the metal preventing them to escape the metal: this depends only on the energy of the electron, which does not depend on the intensity of the light, but only on its frequency.
Answer:
43.16°
Explanation:
λ = Wavelength = 1.4×10⁻¹⁰ m
θ₁ = 20°
n can be any integer
d = distance between the two slits
Since for the first bright fringe, n₁ = 1
n₂ = 2 for second order line
The relation between the distance of the slits and the angle through which it is passed is:
dsinθ=nλ
As d and λ are constant
∴ Angle by which the second order line appear is 43.16°