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Electrons in an electron microscope have kinetic energy of 2.0 * 10 eV. (Assume non-relativistic kinetic energy 1/2 mv) (a) Find the de Broglie wavelength of the electrons: (b) Find the ratio of this wavelength to the wavelength of light at the middle of the visible spectrum (550 nm): (c) How many times greater magnification is theoretically possible with this microscope than with a light microscope? Find the de Broglie wavelength of an electron accelerated across a 32,160 potential difference in a TV tube. Is it reasonable to treat the electron beam as a beam of particles?

Transcript

00:01 In the question first let us note on the given quantities the kinetic energy k is given as 2 .0 into 10 to the power for electron volts the mass of electron ame is given by 9 .1 into 10 to 3 1 kgues and the planks constant h is given as 6 .6 to 6 to 6 to 6 minus 34 j.

00:31 So for the first part the formula to calculate the deep replace wavelength is given by lambda d is equal to h by p where p is equals to root over 2 m e multiplied by the kinetic energy or we can write the d wavelength wavelength lambda d is equal to h by root over 2 m e multiplied by the kinetic energy so now first we have to convert the kinetic energy from ev to jules so k equals to 2 .0 into 10 to 3 to minus 16 in electron volts which is equals to 2 .0 into 10 to 3 .5 minus 16 multiplied by 1 .6 into 10 to 3 to 10 to the hour minus 15 jules so now by putting in the formula we get the deprogative wavelength lambda d is equal to 6 .6 to 6 .6 to 6 into 10 to hour minus 34 divided by root over 2 multiplied by 9 .1 into 10 to 2 minus 31 multiplied by 3 .2 into 10 to 2 minus 1 or upon solving we get the d -prope d -waple number d is equal to 6 .626 into 10 to minus 34 divided by 7 .631 into 10 to 3 .23 or the wavelength is equal to 8 .68 into 10 to the power minus 19 meters.

02:38 So this is the dprocary wavelength in the second part we are asked to find out the ratio of the wavelengths to the wavelength of light so the wavelength of light is given as lambda c is supposed to 550 nanometer so the formula to calculate the ratio of wavelength is given by lambda ratio is equal to lambda dibroquely divided by the lambda or wavelength of light so by putting the values we get lambda r a is equal to lambda d we got the value of the deep rogd wavelength is equals to 8 .68 into t into 2 to minus 12 meters so 8 .68 into 10 to 2 to 2 minus 12 divided by 550 in 10 to 2 of minus 12 divided by 550 in 10 to 0 .9 which is equal to around 15 .78 into 10 to 2ba minus 6.

03:55 Therefore the ratio of the wavelength is given as lambda array is equal to 15 .78 into 10 to 2a minus 6...

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