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12345 [234]
3 years ago
14

What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incide

nt light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Physics
1 answer:
Softa [21]3 years ago
3 0

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

<u>c. The incident light must have at least as much energy as the electron work function</u>

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A 160 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm and released. A student with a stopwatch finds
Yanka [14]

Answer:

k = 3.41 N/m

Explanation:

The time period is given as:

T = \frac{time\ taken}{No.\ of\ oscillations} \\\\T = \frac{19\ s}{14} \\\\T = 1.36\ s

Another formula for the time period of the spring-mass system is:

T = 2\pi\sqrt{\frac{m}{k}} \\\\(1.36\ s)^2 = 4\pi^2\frac{0.16\ kg}{k}\\\\k = \frac{(4\pi^2)(0.16\ kg)}{(1.36\ s)^2}\\\\

<u>k = 3.41 N/m</u>

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3 years ago
Which is a correct step in a scientific experiment involving acids and bases?
yuradex [85]

Answer:

b) using an indicator to measure the hydrogen ion concentration of a solution

6 0
4 years ago
A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

\epsilon = -\frac{\Delta \Phi_B}{\Delta t} (1)

where

\Delta \Phi_B is the variation of magnetic flux through the coil

\Delta t = 2.0 ms = 0.002 s is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

\Phi_B = BA

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2

At the beginning, the magnetic field is

B_i = 1.5 T

so the flux is

\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

\Phi_f = 0

so the variation of magnetic flux is

\Delta \Phi = 0-0.068 Wb=-0.068 Wb

And substituting into (1) we find the average emf in the coil

\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

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3 years ago
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Arte-miy333 [17]
The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.

so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0

The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.
3 0
4 years ago
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Oksanka [162]
<span>A(n) alpha particle is a particle that contains two protons, two neutrons, and has a 2 charge.
</span>
b. alpha
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3 years ago
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