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2 years ago

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During the photoelectric effect experiment, a photon is emitted with 5.73 x10-20 J of kinetic energy. If the work function of th

e photosensitive surface is 0.540 eV, which is the frequency of the incident light? Show your work. (h = 6.626 x 10-34 J·s; 1 eV = 1.60 x 10-19 J)
Answer text

1 answer:

Lapatulllka [165]2 years ago

4 0

Answer:

Frequency of incident light = 995×10 12

Explanation:

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Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

alexdok [17]

Explanation :

It is given that,

Length of silicon bar, $l=10\mu m= 0.001\ cm$

Free electron density, $n_e=104\ cm^3$

Hole density, $n_h=1016\ cm^3$

$\mu_n=1200\ cm^2/Vs$

$\mu_p=500\ cm^2/Vs$

The total current flowing in bar is the sum of drift current due to hole and the electrons.

$J=J_e+J_h$

$J=nqE\mu_n+pqE\mu_p$

where, n and p are electron and hole densities.

$J=Eq[n\mu_n+p\mu_p]$

we know that $E=\dfrac{V}{l}$

So, $J=\dfrac{V}{l}q[n\mu_n+p\mu_p]$

$J=\dfrac{1.6\times 10^{-19}\ C}{0.001\ V}[104\ cm^{-3}\times 1200\ cm^2/V\ s+1016\ cm^{-3}\times 500\ cm^2/V\ s]$

$J=1012480\times 10^{-16}\ A/m^2$

or

$J=1.01\times 10^{-9}\ A/m^2$

Current, $I=JA$

A is the area of bar, $A= 20\mu m=0.002\ cm$

So, $I=1.01\times 10^{-9}\ A/m^2\times 0.002\ m^2$

Current, $I=2.02\times 10^{-12}\ A$

<em>So, the current flowing in silicon bar is </em> $I=2.02\times 10^{-12}\ A$ .

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4 years ago

An object weighs 10N in air and 70N in water .What is it's weight when immersed in a liquid of relative density 1.5?

Morgarella [4.7K]

<h3>Answer:</h3>

5.5 N

<h3>Explanation:</h3>

<u>We are given;</u>

Real weight of an object in air as 10 N
Apparent weight in water as 7 N
Relative density of a liquid is 1.5

We need to calculate the apparent weight when in liquid .

First we calculate upthrust

Upthrust = Real weight - Apparent weight

= 10 N - 7 N

= 3 N

Then calculate the upthrust in the liquid.

we need to know that;

Relative density of a liquid = Upthrust in liquid/Upthrust in water

Therefore;

1.5 = U ÷ 3 N

Upthrust in liquid = 3 N × 1.5

= 4.5 N

Therefore, the upthrust of the object in the liquid is 4.5 N

But, Upthrust = Real weight - Apparent weight

Therefore;

Apparent weight in Liquid = Real weight - Upthrust

= 10 N - 4.5 N

= 5.5 N

Thus, the weight when the object is immersed in the liquid is 5.5 N

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3 years ago

The fundamental force underlying all chemical reactions is

ladessa [460]

The fundamental force underlying all chemical reactions is D) electrical

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3 years ago

A boy of mass 80 kg slides down a vertical pole, and a frictional force of 480 N acts on him. What is his acceleration as he sli

olga_2 [115]

Answer:

His acceleration is $\overrightarrow{a}=4\frac{m}{s^{2}}$

Explanation:

Newton's second law states that acceleration of a body is cause by a net force, the relation between them is:

$\sum\overrightarrow{F}=m\overrightarrow{a}$

On the boy there're acting two forces, his weight (W) that points downward and the frictional force (f) that points upward (they boy moves downward and friction always is opposite to movement). So $\sum\overrightarrow{F}=\overrightarrow{W}+\overrightarrow{f}$ so (1) is:

$\overrightarrow{W}+\overrightarrow{f}=m\overrightarrow{a}$

Using the positive direction downward weight and gravitational acceleration(g) are positive and friction force is negative:

$W-f=m\overrightarrow{a}$ , solving for a:

$\overrightarrow{a}=\frac{W-f}{m}$ , weight is mg:

$\overrightarrow{a}=\frac{mg-f}{m}=\overrightarrow{f}=\frac{(80kg)(10\frac{m}{s^{2}})-(480)}{80}$

$\overrightarrow{a}=4\frac{m}{s^{2}}$

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4 years ago

What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

svetoff [14.1K]

Base on my research, within 2 hours you have a number of atoms which remain.
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0

So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0

It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms

N0= 5.55*10*8/0.2057 = 2.698*10^9 atoms

Therefore, 2.698*10^9 atoms is the number of N0

4 0

3 years ago