Question

1. The ground state energy of a harmonic oscillator is 5eV. If the oscillator undergoes a transition from its n=4 to its n=3 level by emitting a photon, what is the energy (in eV) of the emitted photon? 10 2. A very thin sheet of plastic (n=1.60) covers one slit of a double-slit apparatus illuminated by 680−nm light. The center point on the screen, instead of being a brightness maximum, is dark. What is the minimum thickness of the plastic (in nm)? 570 3. A standing wave is described by the wavefunction (SI units) y(x,t)= 6sin(πx/2)cos(100πt). From the equation, directly identify the wavelength and frequency of the wave, respectively, in SI units 4,50 4. A piece of glass has a thin film of gasoline floating on it. A beam of light is shining on the film, perpendicular to it. If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and glass are 1.40 and 1.50, respectively, whatis the minimum nonzero thickness (in nm ) of the film to see a bright reflection? 200 5. What is the lowest possible energy (in eV ) of an electron in hydrogen if its orbital angular momentum is ℏ12

​ ? −0.850 6. Two particles in a high-energy accelerator experiment are approaching each other headon, each with a speed of 0.9520c as measured in the laboratory. What is the magnitude of the velocity of one particle relative to the other? 0.9988c 7. In terms of the ground-state energy, E1,1,1​, what is the energy of the highest level occupied by an electron when 10 electrons are placed into a cubical box (3D box with infinite-potential walls)? 3E1​,1,1 8. After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings (oscillations) in 136 s. What is the value of g( in m/s2 ) on this planet? 10.7

Answer 1

Step-by-step explanation:

1. The energy of the emitted photon can be calculated using the energy difference between the two levels:

E = E_final - E_initial

= 5 eV - E_n=3

= 5 eV - 3 eV

= 2 eV

Therefore, the energy of the emitted photon is 2 eV.

2. The dark center point on the screen indicates destructive interference, which occurs when the path difference between the two slits is an odd multiple of half the wavelength.

The minimum thickness (d) of the plastic can be calculated using the formula for the path difference:

d = (λ / (2 * n)) * (2m - 1)

Given:

λ = 680 nm

n = 1.60

m = 1 (to ensure the center point is dark)

Substituting the values into the formula:

d = (680 nm / (2 * 1.60)) * (2 * 1 - 1)

= (680 nm / 3.20) * 1

= 212.5 nm

Therefore, the minimum thickness of the plastic is approximately 212.5 nm.

3. The given wavefunction is y(x,t) = 6sin(πx/2)cos(100πt).

From the equation, we can directly identify the wavelength (λ) and frequency (f) of the wave:

Wavelength (λ) = 2π / (π/2) = 4

Frequency (f) = 100π

Therefore, the wavelength of the wave is 4 (SI units) and the frequency is 100π (SI units).

4. The minimum nonzero thickness of the film can be calculated using the formula for the minimum thickness required for constructive interference:

t = (m + 1/2) * (λ / (2 * n_glass))

Given:

λ = 560 nm

n_gasoline = 1.40

n_glass = 1.50

m = 0 (to see a bright reflection)

Substituting the values into the formula:

t = (0 + 1/2) * (560 nm / (2 * 1.50))

= 280 nm / 3

= 93.33 nm

Therefore, the minimum nonzero thickness of the film is approximately 93.33 nm.

5. The lowest possible energy of an electron in hydrogen with orbital angular momentum ℓ is given by the equation:

E = -13.6 eV / (n^2)

Given:

ℓ = ℏ / 2 (half of the reduced Planck's constant)

n = 12

Substituting the values into the equation:

E = -13.6 eV / (12^2)

= -13.6 eV / 144

≈ -0.0944 eV

Therefore, the lowest possible energy of the electron in hydrogen is approximately -0.0944 eV.

6. The magnitude of the velocity of one particle relative to the other can be calculated using the relativistic velocity addition formula:

v' = (v1 + v2) / (1 + (v1 * v2) / c^2)

Given:

v1 = v2 = 0.9520c

Substituting the values into the formula:

v' = (0.9520c + 0.9520c) / (1 + (0.9520c * 0.9520c) / c^2)

= 1.9040c / (1 + 0.9520^2)

≈ 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is approximately 0.9988c.

7. In a cubical box, the energy levels are given by the formula:

E_nx,ny,nz = (π^2 * ℏ^2) / (2m) * (nx^2 + ny^2 + nz^2)

Given:

10 electrons placed in the box

To find the highest level occupied by an electron, we need to distribute the electrons into the energy levels starting from the lowest energy level (E_1,1,1).

The highest level occupied by an electron will have nx = ny = nz = 5 (since the total number of electrons is 10, and nx + ny + nz = 10).

Substituting the values into the formula:

E_5,5,5 = (π^2 * ℏ^2) / (2m) * (5^2 + 5^2 + 5^2)

= 3E_1,1,1

Therefore, the energy of the highest level occupied by an electron when 10 electrons are placed into a cubical box is 3 times the energy of the ground-state level (E_1,1,1).

8. The period (T) of a simple pendulum can be calculated using the formula:

T = 2π * √(L / g)

Given:

L = 50.0 cm

Number of complete swings (oscillations) = 100

T = 136 s

The value of g (acceleration due to gravity) can be calculated using the formula:

g = (4π^2 * L) / T^2

Substituting the values into the formula:

g = (4π^2 * 0.5 m) / (136 s)^2

≈ 10.7 m/s^2

Therefore, the value of g on this planet is approximately 10.7 m/s^2.