Answer:
The number of bright-dark fringe is 42
Solution:
As per the question:
Wavelength of light, ![\lambda = 500\ nm = 500\times 10^{- 9}\ m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20500%5C%20nm%20%3D%20500%5Ctimes%2010%5E%7B-%209%7D%5C%20m)
Length of the glass cell, x = 3.73 cm = 0.0373 m
Refractive index, ![\mu = 1.00028](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.00028)
Now,
To calculate the bright-dark fringe shifts, we use the formula given below:
![d_{m} = \frac{2x}{\lambda }\times (\mu - 1)](https://tex.z-dn.net/?f=d_%7Bm%7D%20%3D%20%5Cfrac%7B2x%7D%7B%5Clambda%20%7D%5Ctimes%20%28%5Cmu%20-%201%29)
Now, substituting the appropriate values in the above formula:
![d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)](https://tex.z-dn.net/?f=d_%7Bm%7D%20%3D%20%5Cfrac%7B2%5Ctimes%200.0373%7D%7B500%5Ctimes%2010%5E%7B-%209%7D%7D%5Ctimes%20%281.00028%20-%201%29)
≈ 42
Answer:
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Answer:
7,217*10^28 atoms/m^3
Explanation:
- Metal: Vanadium
- Density: 6.1 g/cm^3
- Molecuar weight: 50,9 g/mol
The Avogadro's Number, 6,022*10^23, is the number of atoms in one mole of any substance. To calculate the number of atoms in one cubic meter of vanadium we write:
1m^3*(100^3 cm^3/1 m^3)*(6,1 g/1 cm^3)*(1 mol/50,9g)*(6,022*10^23 atoms/1 mol)=7,217*10^28 atoms
Therefore, for vanadium we have 7,217*10^28 atoms/m^3
Answer:
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