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deff fn [24]
3 years ago
8

An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can a

bsorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (m el = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, ℏ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s)
Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

The longest wavelength of light is 209 nm.

Explanation:

Given that,

Spring constant = 74 N/m

Mass of electron m= 9.11\times10^{-31}\ kg

Speed of light c= 3\times10^{8}\ m/s

We need to calculate the frequency

Using formula of frequency

f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Where, k= spring constant

m = mass of the particle

Put the value into the formula

f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}

f=1.434\times10^{15}\ Hz

We need to calculate the longest wavelength that the electron  can absorb

\lambda=\dfrac{c}{f}

Where, c = speed of light

f = frequency

Put the value into the formula

\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}

\lambda=2.092\times10^{-7}\ m

\lambda=209\ nm

Hence, The longest wavelength of light is 209 nm.

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Answer:

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Explanation:

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ω² L C =1

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The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a
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Answer: 247.67 V

Explanation:

Given

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Potential at V_c=785\ V

when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1

Change in Kinetic Energy of particle moving under the Potential From C to B

q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2

Divide 1 and 2 we get

\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}

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Answer:

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(Credit: Google)

7 0
3 years ago
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