Assume that the de Brogie wave associated with an electron can can from a standing wave between the atome arrange in a one dimensional array with nodes at each of the atomic sites it is found that one such standing wave if the distance d between the aloms of the arry is 2M139waveisaga∈formadifdis∈creased→2.5Å\displaystyle.{A}{s}{i}{m}{i}{l}{a}{r}{s}{ an{{d}}}\in{g}\in{t}{h}{e}{d}{i}{s}{ an{{c}}}{e}{\quad ext{if}\quad}{d}{f}\in{d}{t}{h}{e}{e}
e{r}{g}{y}{o}{f}{t}{h}{e}{e}\le{c}{t}{r}{o}{n}{s}{v}{e}\lt{s}{\quad ext{and}\quad}{t}{g}{h}{e}\le\ast{v}{a}{l}{u}{e}{o}{f}{d}{f}{\quad ext{or}\quad}{w}{h}{i}{c}{h}{t}{h}{e}{s}{ an{{d}}}\in{g}{w}{a}{v}{e}{t}{y}{p}{e}{d}{e}{s}{c}{r}{i}{b}{e}{d}{a}{b}{o}{v}{e}{c}{a}{n}{\mathfrak{{o}}}{m}.

Question

Assume that the de Brogie wave associated with an electron can can from a standing wave between the atome arrange in a one dimensional array with nodes at each of the atomic sites it is found that one such standing wave if the distance d between the aloms of the arry is 2M139waveisaga∈formadifdis∈creased→2.5Å\displaystyle.{A}{s}{i}{m}{i}{l}{a}{r}{s}{	an{{d}}}\in{g}\in{t}{h}{e}{d}{i}{s}{	an{{c}}}{e}{\quad	ext{if}\quad}{d}{f}\in{d}{t}{h}{e}{e}
e{r}{g}{y}{o}{f}{t}{h}{e}{e}\le{c}{t}{r}{o}{n}{s}{v}{e}\lt{s}{\quad	ext{and}\quad}{t}{g}{h}{e}\le\ast{v}{a}{l}{u}{e}{o}{f}{d}{f}{\quad	ext{or}\quad}{w}{h}{i}{c}{h}{t}{h}{e}{s}{	an{{d}}}\in{g}{w}{a}{v}{e}{t}{y}{p}{e}{d}{e}{s}{c}{r}{i}{b}{e}{d}{a}{b}{o}{v}{e}{c}{a}{n}{\mathfrak{{o}}}{m}.

Filo · Accepted Answer

Answer: ASolution: As nodes are formed at each of the atomic sites , heance

‘2 \overset{˚}{A} = n ((l amb d a) / (2)) ‘

…(i)
[:'

‘ D i s t an ce b e tw ee n s u ccess i v e n o d es ‘

= lambda//2]

‘‘

and 2.5 Å= (n + 1) (lambda)/(2)

‘ : . (2.5) / (2) = (n + 1) / (n), (5) / (4) = (n + 1) / (n) or n = 4‘

Heance , from equatio(1).

‘2.5 \overset{˚}{A} = 4 (l amb d a) / (2) i, e, l amb d a = 1 \overset{˚}{A} ‘

Now , de broglie wavelength is given by

‘ l amb d a = (h) / (s q r t (2 m K)) orK = (h^{(} 2)) / (l amb d a^{(} 2) 2 m) ‘

‘ : . K = ((6.63 xx 1 0^{(} - 34))^{(} 2)) / ((1 xx 1 0^{(} - 10))^{(} 2) xx 2 xx 9.1 xx 1 0^{(} - 31) xx 1.6 xx 1 0^{(} - 10)) e V ‘

‘ = ((6.63))^{(} 2)) / (6 xx 9.1 xx 1.6) xx 1 0^{(} 2) e V = 151 e V ‘

d will be miximum , when

‘ n = 1, d_{(} min) = (l amb d a) / (2) = (1 \overset{˚}{A}) / (2) = 0.5 \overset{˚}{A} ‘

Question Text	Assume that the de Brogie wave associated with an electron can can from a standing wave between the atome arrange in a one dimensional array with nodes at each of the atomic sites it is found that one such standing wave if the distance d between the aloms of the arry is $2 M 139 w a v e i s a g a \in f or m a d if d i s \in c r e a s e d \to$ 2.5Å\displaystyle.{A}{s}{i}{m}{i}{l}{a}{r}{s}{\tan{{d}}}\in{g}\in{t}{h}{e}{d}{i}{s}{\tan{{c}}}{e}{\quad\text{if}\quad}{d}{f}\in{d}{t}{h}{e}{e}\ne{r}{g}{y}{o}{f}{t}{h}{e}{e}\le{c}{t}{r}{o}{n}{s}{v}{e}\lt{s}{\quad\text{and}\quad}{t}{g}{h}{e}\le\ast{v}{a}{l}{u}{e}{o}{f}{d}{f}{\quad\text{or}\quad}{w}{h}{i}{c}{h}{t}{h}{e}{s}{\tan{{d}}}\in{g}{w}{a}{v}{e}{t}{y}{p}{e}{d}{e}{s}{c}{r}{i}{b}{e}{d}{a}{b}{o}{v}{e}{c}{a}{n}{\mathfrak{{o}}}{m}.
Answer Type	Text solution:1
Upvotes	150

Text solutionVerified

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