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3 years ago

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When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:

1 answer:

sergij07 [2.7K]3 years ago

5 0

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

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B.) Write a clear, English sentence which describes a general relationship between

Lynna [10]

If a student took seven science classes, his/her at score would be 1200.

<h3>What is a graph?</h3>

A graph is a relationship in which data is presented on the axes of cartesian coordinates. The presentation of the graph must be such that it is easy to read off the graph.

Now we can see that the graph is a plot of SAT scores versus the number of science classes that students take. In this case, the slope would tell us the score per subject of a student in the SAT.

a) The mathematical model that re[resents this graph is;

y =100x + 500

b) The slope of the graph must be obtained by the use of two points thus;

m = y2 - y1/x2 - x1

m = 600 - 500/1 - 0

m = 100

c) Using the relation;

y = mx + c

m = 100

c = 500

y = 100(7) + 500

y = 1200

Thus if a student took seven science classes, his/her at score would be 1200.

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1 year ago

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N

kotegsom [21]

Answer:

(A) = 34,560 J

(B) Work done during the round trip is not zero, hence the resistive force is not a conservative force.

Explanation:

This question is incomplete, here is the complete question:

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.? (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

SOLUTION:

distance covered by the girl due east (De) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fw) = 3.1 N due west

distance covered by the girl due west (Dw) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fe) = 3.3 N due east

(A) to find the work done by the resistive force during the round trip we have to get the work done by the resistive force due west and add it to the work done by the resistive force due east

work done by the resistive force due west = (Fw.cosθ) x De

where θ is the angle between the displacement and the force. The

displacement is due east while the force is due west, hence θ = 180°

work done (W1) = (3.1 x cos 180) x 5400 = -16,740 j

work done by the resistive force due east = (Fe.cosθ) x Dw

where θ is the angle between the displacement and the force. The

displacement is due west while the force is due east, hence θ = 180°

work done (W2) = (3.3 x cos 180) x 5400 = -17,820 j

Hence work done during the round trip = W1 + W2 = (-16,740 ) + (-17,820)

= 34,560 J

(B) A conservative force is a force in which the work done by the force in moving an object around a closed path is zero. From part (A) above the work done during the round trip is not zero, hence the resistive force is not a conservative force.

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2 years ago

Select ALL of the following statements that provide evidence that there is friction acting on a cart moving along a level track.

Sati [7]

Answer:

1st and 4th one................

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3 years ago

A turntable is spinning on a frictionless axle when a ball of putty is dropped onto it from above. The putty sticks to the turnt

andrezito [222]

Answer:

Yes

Explanation:

Gravity and forces form the axle (external forces) do not cause forces that tend to cause rotation around the turntable's axle.

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3 years ago

When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener

erastova [34]

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

$K_{max} =E_{photon} - \Phi$ (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h= $6.63\times10^{-34}$ ) times the frequency of light ( $\nu$ ) , then (1) is:

$K_{max} =h\nu - \Phi$ (2)

The frequency of an electromagnetic wave is related with the wavelength ( $\lambda$ ) by:

$\nu=\frac{c}{\lambda}$ (3)

with c the velocity of light (c= $3.0\times10^{8}$ )

Using (3) on (2):

$K_{max} =\frac{hc}{\lambda} - \Phi$

Solving for $\Phi$ :

$\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}$

$\Phi=5.23\times10^{-19} J$

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

$2K_{max} =\frac{hc}{\lambda} - \Phi$

Solving for $\lambda$ :

$\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}$

$\lambda=173 nm$

3 0

3 years ago