Question

74. Number of photons of light with a wavelength 2000 Å
that provides 0.5 J energy is nearly nxhc
(h = 6.6 * 10-34 Js)
7.30
(1) 3 * 1017
(3) 5 × 1017
(4) 4 * 1017
(2) 9 x 1017 0.5 = n*
2000

Answer 1

Answer:

5×10^17

Explanation:

See the picture.

Hope it helps.

Answer 2

The number of photons in the incident light is (3) $5X10^{17}$.

Given,

Wavelength, λ=2000 Å

Energy of the electrons, E=0.5 J.

To find,

the number of photons of light.

Solution:

• Energy of a photon is directly proportional to the incident frequency.
• E∝f(frequency).
• $E=hf$.
• $E=\frac{hc}{lambda}$

• where, h-Plank's constant, c-speed of light and λ-wavelength.
• $h=6.626X10^{-34} J-sec$.
• $c=3X10^{8} m/s$.

Energy of n number of photons,

$E=nX\frac{hc}{lambda}.$

As the energy is in joules, the other components will also be needed to be converted into their S.I. units.

The value of the wavelength in S.I. units will be,

$1 m=10^{10} Å\\1 Å=\frac{1}{10^{10}} m\\1 Å=10^{-10} m.$

$2000Å=2000X10^{-10} \\2000Å=2X10^{3} X10^{-10}\\2000Å=2X10^{-7} m.$

The value of wavelength in S.I. units is $2 X10^{-7} m$.

$E=nX\frac{hc}{lambda}.\\n=\frac{lambdaXE}{hc} \\n=\frac{2 X10^{-7} X0.5 }{6.626X10^{-34}X3X10^{8} } \\n=\frac{1X10^{-7}}{19.878X10^{-26}} \\n=\frac{1X10^{-7+26} }{19.878} \\n=0.05X10^{19}\\n=5X10^{17}.$

Hence the number of photons are $5X10^{17}$.

#SPJ2

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