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3 years ago

if a sample that is 4860 years old has 50 radium atoms remaining, how many atoms were in the original sample?

Physics

1 answer:

snow_lady [41]3 years ago

4 0

243000, take the 4,860 and times it by 50

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

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3 years ago

6 latter word and it has a s and a I and it has mass (9.______________liquids, and gases all have mass.)

Firdavs [7]

Answer:

Solids

Explanation:

Solids, liquids, and gases all have mass.

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kherson [118]

can you have all of them as possibilities ???

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Alborosie

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Explanation:

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