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3 years ago

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A diathermy machine, used in physiotherapy, generates electromagnetic radiation that gives the effect of "deep heat" when absorb

ed in tissue. One assigned frequency for diathermy is 915.00 MHz. What is the wavelength of this radiation

2 answers:

olga nikolaevna [1]3 years ago

5 0

Answer:

The wavelength of the given radiation is 0.327 m

Explanation:

The wavelength and the frequency of a wave are related by the formula, given below:

v = fλ

where.

v = speed of wave

f = frequency of wave = 915 MHz = 915000000 Hz

λ = wavelength of the wave = ?

For the electromagnetic radiations we know that:

v = c = speed of light = 3 x 10^8 m/s

Therefore,

c = fλ

λ = c/f

λ = (300000000 m/s)/(915000000 Hz)

<u>λ = 0.327 m</u>

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

The wavelength of this radiation = 0.3279 m

Explanation:

For a wave, the speed (v), frequency (f) and the wavelength (λ) are related through the relation

v = fλ

f is given in the question as 915 MHz = (915 × 10⁶) Hz = (9.15 × 10⁸) Hz

v = speed of light = (3.0 × 10⁸) m/s (since electromagnetic waves/radiation travel with the speed of light)

λ = ?

λ = (v/f)

λ = (3.0 × 10⁸)/(9.15 × 10⁸)

λ = 0.3279 m

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Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

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$v_{e}$ = velocity of an electron

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Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

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$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

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