gneet

Study Companion

For more NEET and AIIMS mcq on Structure of Atom Visit www.gneet.com
Question
Q257) A certain metal when irradiated with light (ν = 3.2 × 1016 Hz) emits photo electrons with twice kinetic energy as did photo electrons when the same metal is irradiated by light (ν = 2.0 × 1016 Hz). Calculate ν0 ofelectron?



SOLUTION

K. E. = h(ν – ν0 )
K.E. of photoelectrons when ν = 3.2 × 1016 Hz
K.E1= h (3.2 × 1016 – ν0)
K E. of photoelectron when ν = 2.0 × 1016 Hz
KE2 = h(2.0 × 1016 – ν0)
According to question KE1 = 2KE2
∴ h(3.2 × 1016 – ν0) = 2h(2.0 × 1016 – ν0)
3.2 × 1016 – ν0 = 4.0 × 1016 – 2ν0
ν0= 4.0 ×1016 – 3.2 × 1016
= 0.8 × 1016 Hz
= 8 × 1015 Hz
Answer:(b)