Physics, 25.06.2020 04:01 vkyles84

The resolving power of a microscope is proportional to the wavelength used. A resolution of 1.0 10-11 m (0.010 nm) would be required in order to "see" an atom. (a) If electrons were used (electron microscope), what minimum kinetic energy would be required for the electrons

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Answer from: flore7488

K = 13448.64eV

Explanation:

(a) In order to calculate the kinetic energy of the electrons, to "see" the atom, you take into account that the wavelength of the electrons must be of the order of the resolution required (0.010nm).

Then, you first calculate, by using the Broglies' relation, the momentum of the electron associated to a wavelength of 0.010nm:

$p=\frac{h}{\lambda}$ (1)

p: momentum of the electron

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 0.010nm

You replace the values of the parameters in the equation (1):

$p=\frac{6.262*10^{-34}Js}{0.010*10^{-9}m}=6.262*10^{-23}kg\frac{m}{s}$

With this values of the momentum of the electron you can calculate the kinetic energy of the electron by using the following formula:

$K=\frac{p^2}{2m}$ (2)

m: mass of the electron = 9.1*10^-31 kg

$K=\frac{(6.262*10^{-23}kgm/s)^2}{2(9.1*10^{-31}kg)}=2.15*10^{-15 }J$

In electron volts you obtain:

$2.15*10^{-15}J*\frac{6.242*10^{18}eV}{1J}=13448.64eV$

The kinetic energy required for the electrons must be, at least, of 13448.64 eV

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

Answer from: Quest

Ithink it's c. i'm sure

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