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Finger [1]
3 years ago
10

In an experiment you measure a first-order red line for Hydrogen at an angle difference of ΔΘ = 22.78o. The diffraction grating

you are using has 5900 lines per cm.
a) What is the wavelength of this light?

b) What is the value of Rydberg's constant for this measurement?
Physics
1 answer:
azamat3 years ago
4 0

Answer:

a) wavelength = 656.3 nm

b)  the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

Explanation:

Given that;

angle of diffraction Θₓ = 22.78°

incident angle Θ₁ = 0

slit separation d  = 5900 lines per cm = 1/5900 cm = 10⁻²/5900 m = 0.01/5900 m

order of diffraction n = 1

wavelength λ = ?

to find the wavelength, we use the expression

λ = d (sinΘ₁ + sinΘₓ) / n

To find the wavelength λ;

λ = 0.01/5900 × (sin0 + sin22.78° )

λ = 6.5626 × 10⁻⁷ m

λ = 656.3 x 10⁻⁹ m

∴ λ = 656.3 nm

b)

According Balnur's  series spectral lines; n₁ = 3, n₂ = 2 and

λ = R [ 1/n₂² - 1/n₁²]

where  R is Rydberg's constant

from λ = R [ 1/n₂² - 1/n₁²]

R = 1/λ [n₂²n₁² / n₁² - n₂²]

R = 10⁹/ 656.3 [ 9 × 4 / 9 - 4 ]

R = 1.097 × 10⁷ m⁻¹

Therefore the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

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6 0
3 years ago
A rifle of mass 2 kg is horizontally suspended by a pair of strings so that recoil can be measured. The rifle fires a bullet of
aliya0001 [1]

Answer: 1m/s

Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.

Mathematically

Mb×Vb = Mg×Vg

Where Mb = mass of bullet = 1/100 = 0.01 kg

Vb = velocity of bullet = 200 m/s

Mg = mass of gun = 2kg

Vg = recoil velocity of gun =?

0.01×200 = 2×Vg

Vg = 0.01×200/2

Vg = 0.01×100

Vg = 1m/s

3 0
3 years ago
The radiation per unit area from the Sun reaching the earth is 1400 W/m2 , approximately the amount of radiative power per unit
natka813 [3]

Answer:

83.2 W/m^2

Explanation:

The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:

P=\sigma A T^4

where

\sigma is the Stefan-Boltzmann constant

A is the surface area

T is the surface temperature

So, we see that the radiation per unit area is proportional to the fourth power of the temperature:

I \propto T^4

So in our problem we can write:

I_1 : T_1^4 = I_2 : T_2^4

where

I_1 = 1400 W/m^2 is the power per unit area of the present sun

T_1 = 5800 K is the temperature of the sun

I_2 is the power per unit area of sun X

T_2 = 2864 K is the temperature of sun X

Solving for I2, we find

I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2

6 0
3 years ago
11. All objects gain or release heat at the same rate. *
aleksandrvk [35]

The statement is false.

5 0
3 years ago
Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a
In-s [12.5K]

Answer:

T = 1.108\,s

Explanation:

The period of a physical pendulum is:

T = \sqrt{\frac{I_{O}}{m\cdot g \cdot L} }

T=2\cdot \pi \sqrt{\frac{\frac{1}{3}\cdot m \cdot L^{2} }{m\cdot g\cdot L} }

T=2\cdot \pi \sqrt{\frac{L }{3\cdot g} }

The length of the leg is approximately the height of the person:

L = 0.915\,m

The period is:

T = 2\cdot \pi \sqrt{\frac{0.915\,m}{3\cdot (9.807\,\frac{m}{s^{2}} )} }

T = 1.108\,s

4 0
3 years ago
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