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3 years ago

Use the following photoelectric graph to answer the following question:

Physics

1 answer:

8090 [49]3 years ago

8 0

Answer:

2.9 E14 Hz

Explanation:

As we know by Einstein's equation that energy incident on the photo sensitive surface will be used by the surface to eject electron out of the surface with some kinetic energy.

This is given by

$E = \phi + KE$

now the threshold frequency is the minimum frequency of the incident photons due to which electrons are ejected out with minimum kinetic energy or least kinetic energy.

So here when KE = 0 in the graph then corresponding to that position the frequency will be given as threshold frequency

so here from graph when KE = 0

$f = 2.9 E14 Hz$

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For how long is the ball in the air?

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$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

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The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
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$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

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Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

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