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2 years ago

The wavelength of an electron whose velocity is 1.7 x 104 m/s and whose mass is 9.1 x 10-28 g is __________ m.

1 answer:

6 0

The formula for DeBroglie wavelength is λ=h/p, where λ is the wavelength, h=6.626*10^-34 m₂ kg/s is the Planck's constant and p is the momentum of the electron. p=m*v, where m is the mass of the electron and v is the velocity of the electron.

First we need to calculate the momentum p:

m=9.1*10^-28 g=9.1*10^-31 kg because we need to plug in mass in kilograms

p=(9.1*10^-31)*(1.7*10^4)=1.547*10^-26 kg m/s

Now we calculate the wavelength:

λ=(6.626*10^-34)/(1.547*10^-26)=4.283*10^-8 m

So the DeBroglie wavelength of the electron is λ=4.283*10^-8 m.

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A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how

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Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.69}{k}$

$k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}$

b) Expression for rate law for first order kinetics is given by:

$A=A_0e^{-kt}$

where,

k = rate constant

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

$A_0$ = initial activity = 600, 000

$A=600000\times e^{-0.346\times 6}$

$A=75258$

Thus there will be 75258 nuclei left at 6 pm.

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2 years ago

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evablogger [386]

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