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Helga [31]
2 years ago
14

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

al are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?
Check all that apply.

A. The work function of the metal decreases.
B. The number of electrons emitted from the metal per second increases.
C. The maximum speed of the emitted electrons increases.
D. The stopping potential increases.
Physics
1 answer:
qwelly [4]2 years ago
6 0

Answer:

B. The number of electrons emitted from the metal per second increases.

Explanation:

Light consists of photons . Energy of each photon depends upon frequency of light . The increase in intensity increases the number of photons . It does not increase energy of photons .

So if a high intensity light falls on a photosensitive plate , each photon ejects one electron . So number of electrons increases if we increase intensity of photon. It does not increase kinetic energy of ejected electrons . Work function depends upon the nature of plate.

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SashulF [63]

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The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

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Length of the current element dl=(0.5\times10^{-3})j

Current in y direction = 5.40 A

Point P located at \vec{r}=(-0.730)i+(0.390)k

The distance is

|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}

|\vec{r}|=0.827\ m

We need to calculate the magnetic field

Using Biot-savart law

B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

We need to calculate the value of \vec{dl}\times\vec{r}

\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

7 0
3 years ago
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