All categories

3 years ago

What is the longest wavelength of radiation that possesses the necessary energy to break the bond 941?

1 answer:

3 0

<span>Energy is calculated by molecule dividing energy by mole by Avogadro's number (6.022*10^23) 941kJ=9.41*10^5 J so energy by molecule E= 9.41*10^5/6.022*10^23=1.563*10^-18 J Wavelength (w) given by E=hc/w where, E = energy h = planks constant (6.6262 x 10-34 JÂ·s) c = speed of light (3 x 10^8 m/s ) So, w= hc/E = (6.6262*10^-34)*(3*10^8) /1.563*10^-18 = 127.2 Nm Longest wavelength of radiation =127.2 Nm</span>

You might be interested in

Why does ice float on water? a.The temperature of ice is lower than the temperature of water b.The temperature of water and ice

pav-90 [236]

The density of ice is less than the density of water. C

7 0

3 years ago

The first person to map the Milky Way Galaxy using radio waves was

Art [367]

Answer:B

Explanation:Grote reber was the first scientist to map the milky way galaxy using radio waves.

7 0

2 years ago

Read 2 more answers

A spherical ball is dropped through a liquid, explain why it reaches terminal velocity.

Alekssandra [29.7K]

Probably because of the drag coefficient and the density of the liquid.

8 0

3 years ago

What is runoff water?

vitfil [10]

<u>Answer:</u> runoff water is water from rain, snow, or other sources, that flows through the land, and is a major component of the water cycle.

6 0

3 years ago

Read 2 more answers

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

2 years ago

Read 2 more answers