PHYSICS COLLEGE

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer 1

Answer:

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Answer 2

Answer:

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

Learn more about energy of photons here:

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The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees

Answers

Answer:

Angle of transmitted ray is $29.14^(o)$

Explanation:

According to snell's law we have

$n_(1)sin(\theta _(i))=n_(2)sin(\theta r)$

Since the incident medium is air thus we have $n_(1)=1$

By definition of refractive index we have

$n=(c)/(v)$

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

$n_(2)=(3* 10^(8))/(2.5* 10^(8))=1.2$

Thus using the calculated values in Snell's law we obtain

$sin(\theta _(r))=(sin(31.3)/(1.2)\n\n\therefore sin(\theta _(r))=0.4329\n\n\theta_(r)=sin^(-1)(0.4329)\n\n\theta _(r)=29.14$

Answer:

Angle made by the transmitted ray = 25.65°

Explanation:

Speed of light in plastic = v = 2.5 × 10⁸ m/s

refractive index of plastic (n₂) / refractive index of air (n₁)

= speed of light in air c / speed of light in plastic v.

⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2

Angle of incidence = 31.3° = i

n₁ sin i = n₂ sin r

⇒ sin r = (1)(0.5195) / 1.2 = 0.4329

⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°

A suspended platform of negligible mass is connected to the floor below by a long vertical spring of force constant 1200 N/m. A circus performer of mass 70 kg falls from rest onto the platform from a height of 5.8 m above it. Find the maximum spring compression

Answers

Answer:

The maximum spring compression = 3.21 m

Explanation:

The height of the circus performer above the platform connected to string material = 5.8 m

Let the maximum compression of the spring from the impact of the circus performer be x.

According to the law of conservation of energy, the difference in potential energy of the circus performer between the initial height and the level at which spring is compressed to is equal to the work done on the spring to compress it by x

Workdone on the spring by the circus performer = (1/2)kx²

where k = spring constant = 1200 N/m

Workdone on the spring by the circus performer = (1/2)(1200)x² = 600x²

The change in potential energy of the circus performer = mg (5.8 + x)

m = mass of the circus performer = 70 kg

g = acceleration due to gravity = 9.8 m/s²

The change in potential energy of the circus performer = (70)(9.8)(5.8 + x) = (3978.8 + 686x)

600x² = 3978.8 + 686x

600x² - 686x - 3978.8 = 0

Solving this quadratic equation

x = 3.21 m or - 2.07 m

Since the negative answer doesn't satisfy the laws of physics, our correct answer is 3.21 m

Hope this Helps!!!

If a photon has a frequency of 5.20 x 10^14 hertz, what is the energy of the photon ? Given : Planck's constant is 6.63 x 10^-34 joule-seconds.

Answers

For this you would use planck's equation.

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14 x 6.63 x 10^-34
= 3.45 x 10^-19 Joules

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = (1)/(2)mv^(2)$

So, $v = \sqrt{(2E)/(m)}$ .... (1)

Now,

$r = (mv)/(Bq)$

Substitute the value of v from equation (1), we get

$r = (√(2mE))/(Bq)$

Let the radius of the alpha particle is r2.

For proton

So, $r_(1) = \frac{\sqrt{2m_(1)E}}{Bq_(1)}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_(2) = \frac{\sqrt{2m_(2)E}}{Bq_(2)}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$(r_(1))/(r_(2))=(q_(2))/(q_(1))* \sqrt{(m_(1))/(m_(2))}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$(r_(1))/(r_(2))=\frac{2q}}{q}}* \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

Answer:15.95 cm

Explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle $=6.64* 10^(-27) kg$

mass of proton $=1.67* 10^(-27) kg$

charge on alpha particle is twice of proton

radius of Proton is given by

$r=(mv)/(|q|B)$

and Kinetic energy $K=(P^2)/(2m)$

where P=momentum

$P=√(2Km)$

$r=(√(2km))/(qB)$ ---1

Radius for Alpha particle is

$r_(alpha)=\frac{\sqrt{2k\cdot m_(alpha)}}{2qB}$ -----2

Divide 1 & 2 we get

$(r)/(r_(alpha))=(√(m))/(q)* \frac{2q}{\sqrt{m_(alpha)}}$

$r_(alpha)=\sqrt{(6.64* 10^(-27))/(1.67* 10^(-27))}* 0.5$

$r_(alpha)=0.997* 16$

$r_(alpha)=15.95 cm$

Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. This changes the mass that oscillates (more water means more mass) but not the restoring force, which is determined by the stiffness of the glass itself. If you need to raise the frequency of a par- ticular glass, should you add water or remove water?

Answers

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

$f=(1)/(2\pi )\sqrt{(k)/(m)}$

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency

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