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A beam of 450 nm light is incident on a metal having work function 2.0 eV and placed in a magnetic field B. The most energetic electrons emitted perpendicular to the field are bent in circular arcs of radius 20 cm. Find the value of B. |
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Answer» Solution : The KINETIC energy of the most energetic electrons is ` K= (HC/ lambda)- varphi` ` =( 1242eV nm / 450 nm) - 2.0e V.` ` = 0.76 e V = 1.2 XX (10^-19)J` ` The linear momentum = mv = (sqrt(2mK)).` ` = (sqrt (2 xx (9.1 xx (10^-31) kg) xx (1.2 xx (10^-19)J)))` ` = 4.67 xx (10^-25) kg (ms^-1).` ` When a charged particle is SENT perpendicular to magnetic field, it goes along a circle of RADIUS. ` r = (mv/ qB)` ` Thus, 0.20m = (4.67 xx (10^25)kg m (s^-1)/ (1.6 xx (10^-19)C)xx B).` ` or, B= (4.67 xx (10^-25)kg (ms^-1))/((1.6 xx (10^-19)C) xx (0.20m))= 1.46 xx (10^-5).` |
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