A free particle having one electronic charge with initial kinetic energy 9 eV and de Broglie wavelength 1 mm enters a region of V0 potential difference such that new de Broglie wavelength is now 1.5 mm. Then eV0 is

Option 1 - <p>5 eV<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>6 eV<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>13.5 eV<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>15 eV</p>
12 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 1
Detailed Solution:

λ? = h/√2mE? = λ? = h/√2mE?
=> E? = (4/9)E? = 4eV
E? = E? - eV? => V? = 5V

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physics ncert solutions class 11th 2023

physics ncert solutions class 11th 2023

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