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dalvyx [7]
3 years ago
5

Determine the wavelength of incident electromagnetic radiation required to cause an electron transition from the n = 5 to the n

= 7 level in a hydrogen atom.
Physics
1 answer:
ollegr [7]3 years ago
8 0
The correct answer is: wavelength = 4562 nm

Explanation:

Rydberg's formula is given as:
\frac{1}{\lambda} = R[ \frac{1}{n_1^2}  - \frac{1}{n_2^2} ] --- (1)

Where 
R = Rydberg's constant = 1.096 * 10^7 per meter
n_1 = 5
n_2 = 7

λ = Wavelength

Plug in the values in (1):

(1)=> \frac{1}{\lambda} = (1.096 * 10^7)[ \frac{1}{5^2} - \frac{1}{7^2} ]

\frac{1}{\lambda} = (1.096 * 10^7)[ 0.04 - 0.020 ] \\ \lambda =  \frac{1}{(1.096 * 10^7)[0.020 ]} \\ \lambda = 4562 nm
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Why is accelration negative when velocity is positive.​
Sindrei [870]

Answer:

An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

Explanation:

8 0
2 years ago
Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where
fiasKO [112]

The emf induced in the second coil is given by:

V = -M(di/dt)

V = emf, M = mutual indutance, di/dt = change of current in the first coil over time

The current in the first coil is given by:

i = i₀e^{-at}

i₀ = 5.0A, a = 2.0×10³s⁻¹

i = 5.0e^(-2.0×10³t)

Calculate di/dt by differentiating i with respect to t.

di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

V = 43 volts

3 0
3 years ago
An object is moving along a straight line at a
Dominik [7]
The answer for this question is 5 m
6 0
1 year ago
the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su
CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0
2 years ago
What’s the answer???
inna [77]

Answer: ( 2nd ) ( 3 )

Explanation:

4 0
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