JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 7)
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 -
25 keV
500 keV
100 keV
1 keV
Explanation
$$\lambda $$ = $${h \over p}$$ {$$\lambda $$ = 7.5 $$ \times $$ 10$$-$$12}
P = $${h \over \lambda }$$
KE = $${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$$
$$ = {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}$$ J
KE = 25 Kev
P = $${h \over \lambda }$$
KE = $${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$$
$$ = {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}$$ J
KE = 25 Kev
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