2. A photon of wavelength for 4 x10 -7 m strikes on the metal surface the work function of the metal being 2.13 eV. Calculate (i)the energy of the photon in
eV(ii) the kinetic energy of the emission(iii) velocity of the photoelectron
[1ev=1.602x10-19 J]

See explanation

Explanation:

Energy of the photon;

E = hc/λ

h= Planks constant

c= speed of light

λ = wavelength

E= 6.6 × 10^-34 × 3 × 10^8/4 x10 -7

E = 4.95 × 10^-19 J

If 1ev=1.602x10-19 J

x = 4.95 × 10^-19 J

x= 3.1 ev

From Einstein's photoelectric equation;

KE = E - Wo

Where;

KE = kinetic energy of ejected photoelectron

E= energy of the photon

Wo= work function of the metal

KE = 3.1 eV - 2.13 eV

KE= 0.97 eV

KE = 0.97 eV × 1.602x10-19 J

KE = 1.55 × 10^-19 J

KE = 1/2 mv^2

1.55 × 10^-19= 1/2 × 9.1 × 10^-31 × v^2

v= √2 × 1.55 × 10^-19/9.1 × 10^-31

v= 5.8 × 10^5 m/s

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.