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Luba_88 [7]
3 years ago
15

An Argon laser gives off pulses of green light (wavelength = 514 nm). If a single pulse from the laser has a total energy of 10.

0 mJ how many photons are in the pulse?
Chemistry
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

n=2.59\times 10^{16} photons

Explanation:

E=n\times \frac{h\times c}{\lambda}

Where,  

n is the number of photons

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light

Given that, wavelength = 514 nm = 514\times 10^{-9}\ m

Energy = 10.0 mJ = 0.01 J ( 1 mJ = 0.001 J )

Applying the values as:-

0.01=n\times \frac{6.626\times 10^{-34}\times 3\times 10^8}{514\times 10^{-9}}

\frac{19.878n}{10^{17}\times \:514}=0.01

n=2.59\times 10^{16} photons

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An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for
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Answer: The freezing point and boiling point of the solution are -6.6^0C and 101.8^0C respectively.

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Depression in freezing point:

T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = freezing point of solution = ?

T^o_f = freezing point of water = 0^0C

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Now put all the given values in the above formula, we get:

(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_f=-6.6^0C

Therefore,the freezing point of the solution is -6.6^0C

Elevation in boiling point :

T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^0C

k_b = boiling point constant of water = 0.52^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

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w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

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Now put all the given values in the above formula, we get:

(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_b=101.8^0C

Thus the boiling point of the solution is 101.8^0C

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