Suppose you need to image the structure of a virus with a diameter of 50 nm. For

Question

Suppose you need to image the structure of a virus with a diameter of 50 nm. For a sharp image, the wavelength of the probing wave must be 5.0 nm or less. We have seen that, for imaging such small objects, this short wavelength is obtained by using an electron beam in an electron microscope. Why don't we simply use short-wavelength electromagnetic waves? There's a problem with this approach: As the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. Let's compare the energy of a photon and an electron that can provide the same resolution.

Part A

For light of wavelength 4.0 nm , what is the energy (in eV) of a single photon? In what part of the electromagnetic spectrum is this?

Express your answer using two significant figures.

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Part B

For an electron with a de Broglie wavelength of 4.0 nm , what is the kinetic energy (in eV)?

Express your answer using two significant figures.

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Suppose you need to image the structure of a virus with a diameter of 50 nm. For a sharp image, the wavelength of the probing wave must be 5.0 nm or less. We have seen that, for imaging such small objects, this short wavelength is obtained by using an electron beam in an electron microscope. Why don't we simply use short-wavelength electromagnetic waves? There's a problem with this approach: As the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. Let's compare the energy of a photon and an electron that can provide the same resolution.

Part A

For light of wavelength 4.0 nm , what is the energy (in eV) of a single photon? In what part of the electromagnetic spectrum is this?

Express your answer using two significant figures.

E =   eV  

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Incorrect; Try Again; 5 attempts remaining

Part B

For an electron with a de Broglie wavelength of 4.0 nm , what is the kinetic energy (in eV)?

Express your answer using two significant figures.

E =   eV  

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Incorrect; Try Again; 8 attempts remaining

Explanation / Answer

a)

wavelength, lambda=4nm

E=h*C/lambda

E=6.626*10^-34*3*10^8/(4*10^-9)

E=4.97*10^-17 J

E=310.6 ev

( it comes under X-ray)

b)

wavelength of electron, lambda=4nm

p=h/lambda

P=(6.63*10^-34)/(4*10^-9)

P=1.66*10^-25 kg*m/sec

and

K.E=p^2/2m

K.E=(1.66*10^-25)^2/(2*9.1*10^-31)

K.E=1.514*10^-20 J

K.E=0.0945 ev

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