Give the postulates of Planck's Quantum Theory
Answers
Explanation:
Answer:
Planck Eric Weisstein's World of Biography postulated that the energy of oscillators in a blackbody is quantized by
(1)
where n = 1, 2, 3, ..., h is Planck's constant, and is the frequency, and used this postulate in his derivation of the Planck law of blackbody radiation. In fact, electromagnetic radiation is itself quantized, coming in packets known as photons and having energy
(2)
In the other hand, the energy of state n of quantum mechanical simple harmonic oscillator is actually given by the slightly modified form
(3)
Photon, Planck Law, Simple Harmonic Oscillator--Quantum Mechanical
I hope my answer is correct.
Explanation:
To find : The distance between the school and his house
Solution :
Speed of a student = 2 ½ km/h = 5/2km/h
★ A student reaches his school 6 minutes late.
Consider the time be x
60min = 1 hour
Time = (x + 6) min = (x + 6/60) = (x + 1/10)h
As we know that
★ Speed = distance/time
Consider the distance be y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ x + \dfrac{1}{10}}⟹25=x+101y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ \dfrac{10x + 1}{10} }⟹25=1010x+1y
\implies \sf \dfrac{5}{2} = y \times \dfrac{10}{10x +1}⟹25=y×10x+110
\implies \sf \dfrac{5}{2} = \dfrac{10y}{10x +1}⟹25=10x+110y
\implies \sf 5(10x +1) = 20y⟹5(10x+1)=20y
\implies \sf 50x +5=20y⟹50x+5=20y
\implies \sf 50x - 20y = - 5 \: \: \: \bf(Equation \: 1)⟹50x−20y=−5(Equation1)
★ Next day starting at the same time, he increases his speed by 1 km/hour and reaches 6 minutes early.
\implies \sf \dfrac{5}{2} + 1 = \dfrac{y}{ x - \dfrac{1}{10}}⟹25+1=x−101y
\implies \sf \dfrac{5 + 2}{2} = \dfrac{y}{\dfrac{10x - 1}{10}}⟹25+2=1010x−1y
\implies \sf \dfrac{7}{2} = \dfrac{10y}{10x - 1}⟹27=10x−110y
\implies \sf 7(10x - 1) = 20y⟹7(10x−1)=20y
\implies \sf 70x - 20y = 7 \: \: \: \bf(Equation \: 2)⟹70x−20y=7(Equation2)
Subtract both the equations
→ 50x - 20y - (70x - 20y) = -5 - 7
→ 50x - 20y - 70x + 20y = - 12
→ - 20x = - 12
→ x = 12/20 = 3/5
Put the value of x in eqⁿ (1)
→ 50x - 20y = - 5
→ 50 × 3/5 - 20y = - 5
→ 30 - 20y = - 5
→ 30 + 5 = 20y
→ 35 = 20y
→ y = 35/20 = 7/4
→ y = 1 3/4 km
•°• The distance between school and house is 1 3/4 km.