What is the temperature if the peak of a blackbody spectrum is at 18-nm?

Physics

1 answer:

finlep [7]3 years ago

4 0

Answer:

Explanation:

According to Wein's displacement law

$\lambda_{max} T = b$

Where b = 2.898 x 10⁻³ m K

Given ,

$\lambda{max} = 18\times10^{-9}$

T = 161 X 10³ K.

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Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608\times 10^{-9}\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=\dfrac{mD\lambda}{a}$

where

a = width of the slit

$a=\dfrac{mD\lambda}{y}$

$a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}$

a = 0.000167 m

$a=1.67\times 10^{-4}\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

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A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

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