To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a protein molecule that is 2.40 nm in diameter?

Take the mass of an electron to be #9.11x 10^-31# kg.

Answer 1

#sf(4.18xx10^(-20)color(white)(x)J)#

I'll adjust the electron's wavelength so that it matches the protein molecule's diameter:

#sf(lambda=2.4xx10^(-9)color(white)(x)m)#

Using the de Broglie expression, we obtain:

#sf(lambda=h/(mv))#
#:.##sf(v=h/(mlambda))#
#sf(v=(6.63xx10^(-34))/(9.11xx10^(-31)xx2.4xx10^(-9))color(white)(x)"m/s")#
#sf(v=3.03xx10^(5)color(white)(x)"m/s")#
Kinetic energy #sf(=1/2mv^2)#
#:.##sf(KE=1/2xx9.11xx10^(-31)xx(3.03xx10^5)^2color(white)(x)J)#
#sf(KE=4.18xx10^(-20)color(white)(x)J)#
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Answer 2

The kinetic energy of the electrons needed to resolve a protein molecule with a diameter of 2.40 nm in an electron microscope can be calculated using the de Broglie wavelength equation. The kinetic energy required is approximately 4.98 × 10^−18 joules, or 3.11 electron volts (eV).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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