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3 years ago

What is the longest wavelength of light that will emit electrons from a metal whose work function is?

1 answer:

6 0

Best Answer:<span> </span><span>hf = work function + KE
However if you are looking at the max wavelength (or threshold frequency) then there will only be just enough energy for the photoelectrons to be liberated, hence their KE will be 0.
So hf = work function
convert eV to joules, 2.4 x (1.6 x 10^-19) = 3.84 x 10^-19
therefore, hf = 3.84 x 10^-19
f = 3.84 x 10^-19 / planck's constant which is 6.63 x 10^-34
f = 5.79 x 10^14 Hz
c = frequency x wavelength,
wavelength = speed of light/frequency
= (3x10^8)/(5.79x10^14)
=5.18 x 10^-7 metres</span>

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Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.

lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, $q=3\times 10^{-9}\ C$

Distance of the given point from the point charge is, $d=3\ cm=0.03\ m$

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

$E=\frac{kq}{d^2}$

Plug in $k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m$ . Solve for 'E'.

$E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C$

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0

3 years ago

What can be inferred when an atom is electrically charged?

Elena-2011 [213]

When an atom becomes electrically charged the number of electrons or protons stops and they are not equal again. The "extra" electron or proton is not balanced by something inside the atom any longer and it starts attracting itself to othet protons or electrons in other atoms.

<h3>What is atomic structure?</h3>

An atomic structure comprises of positively charged nucleus which is surrounded by negatively charged particles called electron and neutron which is neutral charged.

Unlike charges attract each other while like charges repel each other.

Therefore, When an electron is fully charged, the number of electrons will stop to be unequal again.

Learn more about Atomic charge here.

brainly.com/question/18102056

7 0

2 years ago

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

5 0

2 years ago

Gravitational potential energy is due to....

mote1985 [20]

Answer:

Explanation:

I only think its A because of the gravity part...sorry im not good at explaining

5 0

2 years ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0

3 years ago