The threshold frequency for metallic potassium is 5.46×1014s-1 calculate the maximum kinetic energy and velocity that the emitted electron has the wave length of light shining on the potassium surface is 350nm(Me=9.11×1031kg)

The kinetic energy of the electrons is 2.1  × 10^-19 J while the velocity of the electron is 6.78 × 10^5 m/s.

The work function of the metal is obtained as follows;

Wo = hfo

Wo = work function

h = Plank's constant

fo =  threshold frequency

Wo = 6.63 × 10^-34 Js ×  5.46 × 10^14 s-1 = 3.6 ×  × 10^-19 J

The energy of the incident photon is obtained from;

E = hc/λ

E = energy of incident photon

h = Plank's constant

c = speed of light

λ = wavelength of photon

E = 6.63 × 10^-34 Js × 3 × 10^8/350 × 10^-9

E = 5.7  × 10^-19 J

From the Einstein equation;

KE = E - Wo

KE = 5.7  × 10^-19 J -  3.6 ×  10^-19 J = 2.1  × 10^-19 J

Since;

KE = 1/2mv^2

v = √2KE/m

v = √2 × 2.1  × 10^-19 J/9.11 × 10^-31kg

v = 6.78 × 10^5 m/s

Learn more: link

Given :

The threshold frequency for metallic potassium is 5.46×1014 s⁻¹.

To Find :

The maximum kinetic energy and velocity that the emitted electron has the wave length of light shining on the potassium surface is 350 nm.

Solution :

We know, Maximum kinetic energy is given by :

Hence, this is the required solution.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

the answer is a and have a great day