एक प्रोटॉन, न्यूट्रॉन, इलेक्ट्रॉन तथा `alpha-` कण की ऊर्जाये समान है, तब इनकी दे-ब्रोग्ली तरंगदैर्ध्य की तुलनात्मक

Options

A. `lambda _(p) = lambda _(n) gt lambda _(e) gt lambda_(alpha)`
B. `lambda _(alpha ) = lambda _(p) gt lambda _(n) gt lambda_(e)`
C. `lambda _(e ) = lambda _(p) gt lambda _(n) gt lambda_(alpha)`
D. `lambda _(e ) = lambda _(p) gt lambda _(n) gt lambda_(alpha)`

Answer

B

Explanation

कण की गतिज ऊर्जा `K = (1)/(2) mv^(2) ` या `m v = sqrt(2 m K)`
दे-ब्रोग्ली तरंगदैर्ध्य `lambda = (h)/(mv) = (h)/(sqrt(2 m K))`
`:. K, lambda prop 1 // sqrt(m)`
`:. lambda_(p) : lambda_(n) : lambda_(e) : lambda_(alpha) = (1)/(sqrt(m_(p))) : (1)/(sqrt(m_(n))): (1)/(sqrt(m_(e))) : (1)/(sqrt(m_(alpha)))`
चूँकि `m_(p) = m_(n), lambda_(p) = lambda_(n)`
`m_(alpha) gt m_(p)` अतः `lambda_(alpha) gt lambda_(p), m_(e) lt m_(n)` अतः `lambda_(e) gt lambda_(n), lambda_(alpha) lt lambda_(p) = lambda_(n) lt lambda_(e)`