I found an answer from www.britannica.com

nuclear fusion | Development, Processes, Equations, & Facts ...

Nuclear fusion, process by which nuclear reactions between **light** elements form
heavier elements. ... The vast **energy** potential of nuclear fusion was first
exploited in ... In cases where the interacting nuclei belong to elements **with** low
atomic ... of mass **M**. The formula is B = (Zmp + Nmn − **M**)c^{2}, where **m**p and **m**n
are the ...

For more information, see nuclear fusion | Development, Processes, Equations, & Facts ...

I found an answer from www.khanacademy.org

De Broglie wavelength (video) | Khan Academy

Bohr **model energy** levels (derivation using **physics**) ... **Absorption** and **emission**
... that **a** microscope using **electron** "**matter waves**" instead of **photon light waves**
... or in **a** situation such as the **photoelectric effect**, then I see it as **a** small particle..
. ... As large **number** of **electrons absorb energy**, **energy absorbed per electron** ...

For more information, see De Broglie wavelength (video) | Khan Academy

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The Physics of Subatomic Particles

dimension **5** ems . by **2** ems., and finally hit a fluorescent screen on which ... this
indicates an atomic diameter of about 3.4 x **10**''' **m** for an argon atom, ... quickly
lose their **energy** and **fall** into the nucleus, which was obviously not what wa s ...
Classical physics stated that the emission and **absorption** of **light** and other energ
-.

For more information, see The Physics of Subatomic Particles

I found an answer from web.stanford.edu

Problem Solutions

P is the power density (**W m**−**2**) and f is the frequency in Hz. ... The **absorbed**
photons represent 0.794 of the total **energy**. If this were ... **10 cm** by **10 cm**
**photocell** (**with** 100% quantum efficiency) illu- ... nor load resistance, Iν = **5** A. The
open-circuit voltage is ... The total **light** power that **falls** on the solar image is 4410
**W**. d.

For more information, see Problem Solutions

Given that

Light intensity I = 10^{-5} W m^{-2}

Surface area of sodium photo-cell A = 2 cm^2 = 2 * 10^{-4} m^2

Number of layers n = 5

Work function of the metal \phi = 2 eV

Step 1: Finding the number of conduction electrons

Effective atomic area of a sodium atom A_e = 10^{-20} m^2

Number of conduction electrons n_e = n * \frac{A}{A_e}

n_e = 5 * \frac{2 * 10^{-4}}{10^{-20}}

n_e = 10^{17}

Step 2: The amount of energy consumed per second of the electron

Power of the incident light P = I * A

P = 10^{-5} * 2 * 10^{-4}

P = 2* 10^{-9} W

Relation between power and energy of photon

P = nE

Number of photons per second E=\frac{P}{n_e}

E = \frac{2* 10^{-9}}{10^{17}}

E = 2* 10^{-26} J/s

Step 3: Time needed for photoelectric emissions

Time t = \frac{\phi}{E}

t = \frac{2 ev}{2* 10^{-26}} = \frac{2 * 1.6 * 10^{-19}}{2* 10^{-26}}

t =1.67* 10^{7} s

t = 0.507 years

The time needed for photoelectric emission is almost half a year, which is not realistic. Thus, the description of the wave is in conflict with the experiment.