NEET Questions Solved

NEET - 2011

An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function 2.75 eV. If the stopping potential of the photo-electron is 10V,the value of n is 

(a)3                                    (b)4

(c)5                                    (d)2

E=KEmax+W

=eV0+W

=10+2.75

E=12.75eV

Difference of 4 and 1 energy level is 12.75 eV. So, higher energy level is 4 to ground and excited state is n=3.

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