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3 years ago

7

What kind of radiation is stopped by a piece of paper

2 answers:

lesya692 [45]3 years ago

8 0

Radiation alpha is stopped by paper.

Zina [86]3 years ago

4 0

Three types of radioation - Alpha, Beta, Gamma. hope this helps

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Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

What tripositive ion has the electron configuration [kr] 4d3 ? what neutral atom has the electron configuration [kr] 5s 2 4d2 ?

tigry1 [53]

(a)

Electronic configuration is given as follows:

$[Kr]4d^{3}$

Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.

Thus, electronic configuration of neutral atom is $[Kr]4d^{5}5s^{1}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be $Mo^{3+}$ .

(b) The given electronic configuration is $[Kr]5s^{2}4d^{2}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

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