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Sonbull [250]
3 years ago
11

Find the average energy E for (a) An n-state system, in which a given state can have energy 0,,2,...,n . (b) A harmonic oscillat

or, in which a given state can have energy 0,,2,...(i.e., with no upper limit).
Physics
1 answer:
kicyunya [14]3 years ago
4 0

Answer:

a♦1  E_average = n E₀ / 2 , b) E_average= infinity

Explanation:

The energy values ​​form an arithmetic series, whose sum is

          S = n (a₁ + aₙ) / 2 = n (2a₁ + (n-1) r)/ 2

Where n is the number of terms, a₁ is the first term, aₙ the last term and r is the difference between two consecutive numbers in the series

          r = 2E₀ - 0 = 2E₀

Therefore the sum is

       S = n (0 + n E₀) / 2

      S = n² E₀ / 2

     

The average value is

         E_average = S / n

         E_average = n E₀ / 2

b) the case of harmonic oscillation

We have two possibilities.

- if we take a finite number and terms gives the same previous value

- If we take an infinite number of fears the series gives infinity and the average is also infinite

          E_average= infinity

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\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

{\bold{\blue{GIVEN}}}

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

{ \bold{\green{To  \: Find}}}

REAL DEPTH OF THE OBJECT.

{\red{FORMULA   \:  \: USED }}

Reflective  \:  Index  =  \frac{Real \:  Depth}{Apparent  \: Depth }

\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

Refractive Index = 1.3

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Reflective  \:  Index  =  \frac{Real \:  Depth}{Apparent  \: Depth }  \\  \\ 1.3 =  \frac{Real \:  Depth}{7.7 \: cm}  \\ \\  1.3 \times 7.7 = Real \:  Depth \\  \\ 10.01 \:  \: cm = Real \:  Depth

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