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3 years ago

7

gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are

the wavelengths of these photons

1 answer:

AveGali [126]3 years ago

7 0

Answer:

$\lambda=1.03\times 10^{-3}\ m$

Explanation:

Given that,

The energy of the microwave oven is $1.2\times 10^{-3}\ eV$ .

We need to find the wavelength of these photons.

$1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J$

The energy of a wave is given by :

$E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}$

Put all the values,

$\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m$

So, the wavelength of these photon is $1.03\times 10^{-3}\ m$ .

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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of

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$2.27\cdot 10^{49}$

Explanation:

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where

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Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

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r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

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3 years ago

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

a)

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

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3 years ago