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Vedmedyk [2.9K]
3 years ago
15

What is the length of a one-dimensional box in which an electron in the n=1 state has the same energy as a photon with a wavelen

gth of 800 nm ?
Physics
1 answer:
Pie3 years ago
3 0

Answer:

Length will be 0.491 nm

Explanation:

We have given wavelength of the photon \lambda =800nm=800\times 10^{-9}m

Plank's constant h=6.6\times 10^{-34}J-s

We know that energy of the photon is given by

E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{800\times \times 10^{-9}}=2.475\times 10^{-19}J

We know that energy of photon is also given by

E=\frac{n^2h^2}{8mL^2}=\frac{h^2}{8mL^2}

2.475\times 10^{-19}=\frac{(6.6\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times L^2}

L=0.491\times 10^{-9}m

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270/70^2  = x/80^2

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x = 270(6400)/4900

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Hope this helps


4 0
3 years ago
The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


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Explanation:

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