Answer :

(a) maximum energy (K. Emax) of emitted electron is given by the following equation


K. Emax = hν – W0 ……… equation no. 1


Where h = plank’s constant = 6.63 × 10-34 J. sec


ν = max. frequency of X-ray


W0 = work function (minimum energy required to emit an electron from neutral atom)


W0 = 2.14 eV


And, ν = 6 × 1014 Hz


From equation No. 1



KEmax = 0.34 eV


Or in Joules,


KEmax = 0.34 × 1.6 × 10-19 = 0.546 × 10-19 J


maximum KE of emitted electron is 0.546 × 10-19 J


(b) Stopping Potential is calculated by following relation


eV0 = K.Emax


Magnitude of charge of the electron ‘e’ = 1.6 × 10-19 C


V0 = Stopping Potential



V0 = 0.34 V


Stopping Potential of emitted electron is 0.34 V


(c) K.Emax = 1/2 mv2max


m = mass of the electron = 9.1 × 10-31 kg


vmax = maximum velocity of electron


v2max = 2 K.Emax / m


vmax = (2 × 0.546 × 10-19 J / 9.1 × 10-31)1/2


= 347 km /sec


Maximum velocity of emitted electron is 347 km / sec.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

If light ofPhysics - Board Papers

A. How does one ePhysics - Board Papers

Explain with the Physics - Board Papers