When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

2 years ago

During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a

Sloan [31]

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a= $\frac{38-18}{3}ft/s^2$

Average acceleration,a= $\frac{20}{3}ft/s^2$

Average acceleration,a= $6.67ft/s^2$

Hence, the average acceleration= $6.67ft/s^2$

5 0

1 year ago

The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp

Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0

1 year ago

Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted

34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

Aluminium (Wo)=4eV

Platinum(Wo) =6.4eV

Cesium (Wo) =2.1eV

Beryllium (Wo) = 5.0eV

Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium

Others have zero kinetic energy since no electron is emitted.

Hence the rank is:

K.E(K) > K.E(Cs) > 0 (others)

6 0

2 years ago