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3 years ago

9

Normal human body temperature is about 37°C. What is this temperature in kelvins? What is the peak wavelength emitted by a perso

n with this temperature? In what part of the spectrum does this lie?

1 answer:

Nataly_w [17]3 years ago

5 0

Answer:

9318.07190069 nm

infrared

Explanation:

b = Wien's displacement constant = $2.89\times 10^{-3}\ mK$

T = Temperature = 37+273.15 K

From Wien's displacement law we have

$\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.89\times 10^{-3}}{37+273.15}\\\Rightarrow \lambda=0.00000931807190069\ m=9318.07190069\ nm$

The wavelength is 9318.07190069 nm

This wavelength lies in the infrared spectrum (700 nm-1 mm)

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Explain how longitudinal waves and transverse waves are similar to each other and different from each other.

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For transverse waves, the waves move in perpendicular direction to the source of vibration.
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3 years ago

A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the

Lapatulllka [165]

Answer:13.2,41.4,82.8 ft

Explanation:

Given Person is 6.60 ft from the left hand side mirror

Since the focal length of plane mirror is infinity therefore image and object are equidistant from mirror

Distance of first image on left mirror is

$d_1=6.6+6.6=13.2 ft$

i.e. image is 13.2 ft away from object

Second image

Now the right mirror forms the image of object at distance of 14.1 ft right from right mirror so its image is formed in left mirror at a distance of 34.8 from left mirror

so image distance from person is

$d_2=34.8+6.6=41.4 ft$

Third image now form image of second image is formed on right mirror at a distance of 55.5 right from right mirror

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8 0

3 years ago

Davisson and Germer performed their experiment with a nickel target for several electron bombarding energies. At what angles wou

pochemuha

Answer:

The angle of diffraction are 67.75 deg and 53.57 deg.

Explanation:

Given:

Davisson and Germer experiment with nickel target for electrons bombarding.

Voltages : $38\ eV$ and $54\ eV$

We have to find the angles that is $\phi_3_8$ and $\phi_5_4$ .

Concept:

Davison Germer experiment is based on de Broglie hypothesis where it says matter has both wave and particle nature.
When electrons get reflected from the surface of a metal target with an atomic spacing of $D$ , they form diffraction patterns.
The positions of diffraction maxima are given by $Dsin(\phi) = n\lambda$ .
An atomic spacing is $D = 0.215\ nm$ , when the principal maximum corresponds to n=1
The wavelength is $\lambda$ , and $\lambda =\frac{1.227 \sqrt{V} } {\sqrt{V_o}}\ nm$ .

Solution:

Finding the wavelength at $V_o=38\ eV$ .

⇒ $\lambda_3_8 =\frac{1.227 \sqrt{V} } {\sqrt{38V}}\ nm$

⇒ $\lambda_3_8 =0.199$ nm

Plugging the values of wavelength.

⇒ $sin(\phi)=\frac{\lambda}{D}$

⇒ $\phi_3_8=sin^-1(\frac{0.199}{0.215} )$

⇒ $\phi_3_8 =67.75$ degrees.

Now

For for the electrons with energy $54\ eV$ , $V_0=54V$ the wavelength is.

⇒ $\lambda_5_4 =\frac{1.227 \sqrt{V} } {\sqrt{54V}} = 0.173$ nm

And

⇒ $\phi_5_4=sin^-1(\frac{0.173}{0.215} ) = 53.57$ degrees.

So,

The angles of diffraction maxima are 67.75 deg and 53.57 deg.

6 0

3 years ago