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gayaneshka [121]
4 years ago
6

A simple cyanine dye molecule has a long central carbon chain, with a total of 8 bonds, between the two nitrogen atoms. One elec

tron per bond is essentially free to move up and down this chain. The potential energy for such an electron can be modeled as a one-dimensional box. The distance between each bond is about 0.14 nm.
What would be the transition corresponding to the lowest energy photon that this molecule could absorb? Would such a photon be in the visible range?
Physics
1 answer:
wariber [46]4 years ago
4 0

Answer:

For one dimensional box

E= n^{2}h^{2}/ 8mL^{2}

For lowest energy transition will be from n =1 to n=2

Hence,

hc/\lambda = 2^{2}h^{2}/ 8mL^{2} - h^{2}/8mL^{2}

h= 6.626 *10-34 J.s

m = 9.31 *10-31 Kg

L= 0.14 * 10-9 m

c= 3 *108 m/s

\lambda = 8*c*m*L^{2} / 3*h

\lambda = 22 nm

Not in visible range

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A girl with a mass of 27 kg is playing on a swing. There are three main forces
N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

  • Mass of the girl, m = 27 kg
  • Speed of the girl, v = 3 m/s
  • Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0
3 years ago
A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. Wh
vesna_86 [32]

Answer:t=0.1205

Explanation:

Given

Refractive index of oil\left ( \mu \right )=1.4

Wavelength of red light =675 mm

Condition for constructive interference is

2\mu _{oil}t=\frac{m\lambda }{2}

t=\frac{m\lambda }{4\mu _{oil}}

For m= 1 i.e. for minimum thickness

t=\frac{1\times 675\times 10^{-9}}{4\times 1.4}

t=0.1205 \mu m

5 0
3 years ago
An object with 5 protons, 5 electrons and 6 neutrons would have what charge
Anon25 [30]

Answer:

0

Explanation:

The object would be neutral. There are equal numbers of protons and electrons, so the positive and negative charges cancel one another.

5 0
4 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
3 years ago
Please help on this one?
valina [46]

I believe that it is d correct me if wrong because the higher the temperature the more active the molecules are gonna be, but the graph does not explicitly state that, so you can say the answer is d (if not the answer is c) (sorry if wrong)

3 0
3 years ago
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