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ivolga24 [154]
3 years ago
9

Electron kinetic energies are often measured in units of electron-volts (1 eV 1.6 x 10-19 J), which is the kinetic energy of an

electron that is accelerated through a 1 volt potential. When an aluminum plate is irradiated with UV light of 253.5 nm wavelength the ejected electrons are observed to have an average kinetic energy of about 0.8 eV. Use these results to determine the electron binding energy (or "work function") o of aluminum (in eV units).
Physics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

4.1 eV

Explanation:

Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J

wavelength, λ = 253.5 nm = 253.5 x 10^-9 m

According to the Einstein energy equation

E = W_{o}+K

Where, E be the energy incident, Wo is the work function and K is the kinetic energy.

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J

So, the work function, Wo = E - K

Wo = 7.85 x 10^-19 - 1.28 x 10^-19

Wo = 6.57 x 10^-19 J

Wo = 4.1 eV

Thus, the work function of the metal is 4.1 eV.

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The mass of a sample of iron is 31.5 g. The volume is 5 cm3. What is the density of iron in g/cm3?
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Answer : 6.3 g/cm3
Step by step explanation:
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Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni
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Answer:

1.04μT

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Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

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I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T

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3 years ago
A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0
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1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

I=\frac{9.0}{2.0}=4.5 A

2)

To solve this problem, we can use again the same formula

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s

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8 0
3 years ago
5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in
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The options are;

a. V2 equals 2V1.

b. V2 equals (V1)/2.

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d. V2 equals (V1)/4.

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Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

m' = ρVA

Thus;

At input;

m'1 = ρ1•V1•A1

At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

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Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

Thus, V2 = 2V1

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