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Anna11 [10]
2 years ago
9

When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

is the energy difference between the two energy levels, in J?
Physics
1 answer:
yarga [219]2 years ago
6 0

Answer:

\Delta E=1.64*10^{-18}J

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

\Delta E=h\nu(1)

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

\nu=\frac{c}{\lambda}(2)

Here c is the speed of light. Replacing (2) in (1):

\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J

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Why in a rollercoaster the height of the tracks is either equal to or less than the first track
umka2103 [35]

Answer:

I really don't know

Explanation:

sorry

6 0
2 years ago
At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr
svp [43]
The acceleration of the particle at time t is:
a(t)=24t^2 ft/s^2
The velocity of the particle at time t is given by the integral of the acceleration a(t):
v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s
and the position of the particle at time t is given by the integral of the velocity v(t):
x(t)=\int v(t) = \int (8t^3)=8  \frac{t^4}{4}=2t^4 ft

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
x(2 s)=2 t^4 = 2 (2s)^4=32 ft
5 0
3 years ago
A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t
Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

3 0
2 years ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
The spring constant for the spring in above Table is 20 N/m.
satela [25.4K]
The answer is 0.025J.

W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J
5 0
3 years ago
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