Question 207: Dual Nature of Matter and Radiation

Question

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -

Accepted Answer

$$\lambda $$ = $${h \over p}$$          {$$\lambda $$ = 7.5 $$ 	imes $$ 10
$$-$$12
}
P = $${h \over \lambda }$$
KE = $${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } ight)}^2}} \over {2m}}$$
$$ = {{\left\{ {{{6.6 	imes {{10}^{ - 34}}} \over {7.5 	imes {{10}^{ - 12}}}}} ight\}} \over {2 	imes 9.1 	imes {{10}^{ - 31}}}}$$ J
KE = 25 Kev

Answer

25 keV

Answer

500 keV

Answer

100 keV

Answer

1 keV

Dual Nature of Matter and Radiation

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -

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