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Question: The number of photons of light of wavelengths \[7000{A^ \circ }\] equivalent to \[1J\;\]are : A) \...

The number of photons of light of wavelengths 7000A∘7000{A^ \circ } equivalent to 1J  1J\;are :
A) 3.52×10−183.52 \times {10^{ - 18}}
B) 3.52×10183.52 \times {10^{18}}
C) 50,00050,000
D) 10,000010,0000

Explanation

Solution

Hint : The energy of a photon is related to its frequency and its wavelength. It is directly proportional to frequency and inversely proportional to wavelength. Planck’s constant is used as the proportionality constant in the formula.

Complete step by step solution:
We have the formula of energy-
E=hcλE = \dfrac{{hc}}{\lambda }
Energy of a single photon when the wavelength of the light is 7000A∘7000{A^ \circ }
E=hc7000×10−10E = \dfrac{{hc}}{{7000 \times {{10}^{ - 10}}}}
Now, we know that
h=6.6×10−34J.sec⁡h = 6.6 \times {10^{ - 34}}J.\sec and c=3×108m/sec⁡c = 3 \times {10^8}m/\sec
Putting the above values in above equation, we get-

E=6.6×10−34×3×1087000×10−10⇒E=2.829×10−19J E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{7000 \times {{10}^{ - 10}}}} \Rightarrow E = 2.829 \times {10^{ - 19}}J

Now, Number of photons in 1J\;$$$$ = \dfrac{{1J}}{E}
So, putting the values in the above equation, we get
Number of photons in 1J\;$$$$ = 3.52 \times {10^{18}}

Thus, option B is correct

Additional Information:
In 1905, Einstein used the discrete nature of light to explain the photoelectric effect. The metal surface is highlighted to demonstrate this effect. If the frequency of light is greater than the cutoff frequencyfcfc, electrons are discarded. If the frequency of light falls below this cutoff frequency fc then no photoelectric electrons are emitted. For many metal surfaces the frequency of blue light is greater than fcfcand the frequency of red light is less than fcfc.
If the red light shines on the surface, no electrons are emitted, no matter the intensity of the light. If blue light shines on the surface, electrons are emitted. The number of electrons emitted depends on the intensity of light. But even though the intensity is reduced to a very low value, electrons are still emitted, at a much lower rate.

The photoelectric effect cannot be understood within the light wave image. In order to remove an electron from the surface of the metal a certain amount of energyϕ\phi called the work function of the metal, must be supplied to this electron. The energy of a light beam in a wave diagram is not dependent on frequency, but only on intensity, which is proportional to the square of the amplitude

Einstein explained the photoelectric effect, stating that an electron can absorb the large amount of energy needed to escape the metal from the EM wave by absorbing only one photon. If this photon has enough energy, the electron is released. The excess energy appears as the kinetic energy of the electron. The maximum kinetic energy of an electron is given by E=hf−ϕE = hf - \phi energy. If the photon does not have enough energy, the electron cannot escape from the metal.

Note: After the collision the trajectory of the photon and electron will be at certain angles with respect to the direction of the incoming photon. If we know the energy and momentum of the incoming photon, and angle the trajectory of the outgoing photon along the direction of the incoming photon, we can determine the energy of the outgoing photon using energy and momentum conservation.
The energy of a photon is
E=hf⇒E=hcλ E = hf \Rightarrow E = \dfrac{{hc}}{\lambda }
If the momentum of the photon is pp

hfc=hλ⇒p=Ec \dfrac{{hf}}{c} = \dfrac{h}{\lambda } \Rightarrow p = \dfrac{E}{c}