Question

Please solve this 
The minimum energy required to remove an electron is 
3.7 \times 10 ^{ - 19}3.7×10−19 
Will photon of frequency 
7.5 \times {10}^{14}7.5×1014 
and 
4.3 \times {10}^{14} hz4.3×1014hz 
trigger photoelectric effect. If Yes calculate the maximum kinetic energy​

Answer 1

We will use the following equation :

eV = hf

Where eV = Ionisation energy

h = planks constant = 6.62 × 10⁻³⁴

f = frequency of light.

f = C / Y

Where C = speed of light = 3.0 × 10⁸ n/ s

Y = Wavelength

Substituting in the equation :

3.31 × 10⁻²⁰ = (6.62 × 10 ⁻³⁴ × 3 × 10⁸ ) / Y

Y = (1.986 × 10⁻²⁵) / (3.31 × 10⁻²⁰) = 6 × 10⁻⁶ M

Answer :

6 × 10⁻⁶ M

Hope helps

Answer 2

Explanation:

The minimum energy required to excite an electron from the ground state of an atom to any excited state is called excitation energy. Thus, the energy required to excite an electron in the ground state to the first excited state is called the first excitation energy and so on. The excitation energy of an energy state is the difference between the energy of the excited state and that of the ground state. For example, for the hydrogen atom, the ground state energy(energy of the first orbit) is E1 = -13.6 eV and energy of the second orbit is E2 = -3.4 eV. This means to excite hydrogen atom, the energy required to be given to it is E2 – E1 = -3.4 + 13.6 = 10.2 eV. Hence, 10.2 eV is the first excitation energy of the hydrogen atom. Similarly, the II, the III excitation energy of hydrogen atom are E3 – E1, E4 – E1 etc.