फोटॉन की दे-ब्रोग्ली तरंगदैर्ध्य,इलेक्ट्रॉन की तरंगदैर्ध्य की दोगुनी है इलेक्ट्रॉन की चाल `v_(e) = (c)/(100)`, तब

Options

A. `(E_(e))/(E_(p)) = 10^(-4)`
B. `(E_(e))/(E_(p))= 10^(-2)`
C. `(P_(e))/(m_(e)C)= 10^(-2)`
D. `(P_(e))/(m_(e)C) = 10^(-4)`

Answer

B, C

Explanation

इलेक्ट्रॉन के लिये, `lambda_(e) = (h)/(m_(e)v_(e)) = (h)/(m_(e) (c//100)) = (100 h)/(m_(e) c)" "` . . . (i)
गतिज ऊर्जा, `E_(e) = (1)/(2) m_(e) v_(e)^(2)` या `m_(e) v_(e) = sqrt(2 E_(e) m_(e))`
`lambda_(e) = (h)/(m_(e) v_(e)) = (h)/(sqrt(2 m_(e) E_(e)))` या `E_(e) = (h_(2))/( lambda_(e)^(2) m_(e))" "` . . . (ii)
फोटॉन की तरंगदैर्ध्य `lambda_(p)`, ऊर्जा `= E_(p) = (hc)/( lambda_(p) = (hc)/(2 lambda_(e)" "` (`:. lambda_(p) = 2 lambda_(e)` )
`:. (E_(p))/( E_(e)) = (hc)/(2 lambda_(e)) xx (2 lambda_(e)^(2) m_(e))/(h_(2)) = (lambda_(e) m_(e) c)/(c) = (100 h)/(m_(e) c) xx (m_(e) c)/(h) = 100`
`:. (E_(e))/(E_(p)) = (1)/(100) = 10^(-2)`
इलेक्ट्रॉन के लिये, `p_(e) = m_(e) v_(e) = m_(e) xx c//100`
`:. (p_(e))/( m_(e) c) = (1)/(100) = 10^(-2)`