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1 year ago

An electon in a box absorbs light. The longest wavelength in the absorbtion spectrum is 700 nm . How long is the box

Physics

1 answer:

vova2212 [387]1 year ago

4 0

Answer:

8.01 (10^-10) m

Explanation:

use the Particle in a box equation

E = (n^2) (h^2) / (8 m (L^2))

and energy of a particle as its wavelength equation

h c / lamda = E

combine the equations to get

h c / lamda = (n^2) (h^2) / (8 m (L^2))

get L alone

L^2 = (2^2 - 1^2) h lamda / 8 m c

the longest wavelength is produced by a jump from the first excited state to the ground so

n = 2 to n = 1

L^2 = (n1^2 - n2^2) h lamda / (8 m c)

h = 6.6(10^-34) J*s

c = 3(10^8) m/s

m = 9(10^-31) kg

lamda = 700(10^-9) m

n1 =2

n2=1

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3 years ago

the amplitude of the wave is which of the following A. Resonance B. Reflection C. Energy D. Magnitude

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8 0

3 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

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3 0

2 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

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BartSMP [9]

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3 years ago