Question

Find energy of the photons which correspond to the light of frequency $3\times {{10}^{15}}Hz$(Hint: $h$ Planck's constant $=6.626\times {{10}^{-34}} Js$?
A. $2.988\times {{10}^{-18}}J$
B. $0.988\times {{10}^{-18}}J$
C. $1.308\times {{10}^{-18}}J$
D. $1.988\times {{10}^{-18}}J$

Answer 1

The energy delivered by atomic combination is passed on from the core of the Sun by light particles and warmth, called photons. When consolidating two protons in a core of deuterium to make a helium core, photons are delivered.

Complete step by step answer:
The recurrence is the quantity of waves that pass a point in space during any time stretch, generally one second. We measure it in units of cycles (waves) every second, or hertz.
The Planck steady (meant h) is a characteristic consistent named after Max Planck, one of the organizers of quantum hypothesis. An intently related amount is the decreased Planck constant Planck found in 1901 is consistent in his investigation of dark body radiation.
According to the formula,
$E=hv$
Where $E$ is energy of photons,
$h$ is planck's constant
$v$ is frequency of photons
Here,

$$v=3\times {{10}^{15}}\,Hz \\\ h=6.626\times {{10}^{-34}} \,Js$$

So,

$$=6.626\times {{10}^{-34}}\times 3\times {{10}^{15}} \\\ =19.878\times {{10}^{-19}}\,J \\\ =1.9878\times {{10}^{-18}}\,J$$

**Hence, the answer is D.$1.9878\times {{10}^{-18}}\,J$

Additional Information: **
Envision a pole of yellow daylight radiating through a window. Quantum material science discloses to us that the pillar is made of zillions of small bundles of light, called photons, spilling through the air. Presently, Polish physicists have made the first historically speaking 3D image of a solitary light molecule.

Note:
The energy of the photon relies upon its recurrence (how quick the electric field and attractive field squirm). The higher the recurrence, the more energy the photon has. As the recurrence of a photon goes up, the frequency goes down, and as the recurrence goes down, the frequency increments.