2 years ago

How close does the proton get to the line of charge?

Physics

1 answer:

loris [4]2 years ago

3 0

Answer:

12 cm

Explanation:

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a. 400 N downward

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In general, metalloids are slightly reactive.

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The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

Verizon [17]

Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

The velocity is $v = 2.371*10^6 \ m/s$

The mass of electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

$E = \frac{h * c}{\lambda}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$

Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

=> $E_k = 2.56 *0^{-18} \ J$

Generally the ionization energy is mathematically represented as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$

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3 years ago