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faust18 [17]
3 years ago
8

White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w

ork function of barium is 2.52 eV, find:
(a) the maximum kinetic energy of electrons ejected from this surface.
(b) the range of frequencies for which no electrons are ejected.
Physics
1 answer:
REY [17]3 years ago
5 0

Answer:

0.7515875 eV

4\times 10^{14}\leq f

Explanation:

f = Maximum frequency = 7.9\times 10^{14}\ Hz

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

W = Work function = 2.52 eV

Converting to Joules

W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J

Maximum photon energy is given by

E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J

Maximum Kinetic energy is given by

K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J

Converting to eV

1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz

The range of frequencies for which no electrons are ejected is

4\times 10^{14}\leq f

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
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20.4 m/s^{2}

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To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

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Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

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Now that we know the net force, plug in the numbers for the formula

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a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

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