JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 17)
A metal surface is illuminated by a radiation of wavelength 4500 $$\mathop A\limits^o $$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$$^\circ$$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :
1.36 eV
1.69 eV
2.78 eV
2.23 eV
Explanation
$${{hc} \over \lambda } - \phi = KE$$ ...... (i)
$$R = {{mv} \over {Bq}} = {{\sqrt {2m(KE)} } \over {Bq}}$$ ...... (ii)
Putting the values,
$$\phi \simeq 1.36$$ eV
Comments (0)
