An electron practically at rest, is initially accelerated through a potential difference of 100 volts. It then has a de Broglie wavelength . It then gets retarded through 19 volts and then has a wavelength . It further retardation through 32 volts changes

Question

An electron practically at rest, is initially accelerated through a potential difference of 100 volts. It then has a de Broglie wavelength $= λ_{1} A^{0}$ . It then gets retarded through 19 volts and then has a wavelength $λ_{2} A^{0}$ . It further retardation through 32 volts changes the wavelength to $λ_{3} A^{0}$ . What is the value of $\frac{λ_{3} - λ_{2}}{λ_{1}}$ ?

Smart Achievers · Accepted Answer

(3)

$λ = \frac{h}{\sqrt{2 m e V}}, λ_{1} = \frac{h}{\sqrt{2 m e (100)}} = \frac{h}{10 \sqrt{2 m e}} = \frac{k}{10}, k = \frac{h}{\sqrt{2 m e}}$

$λ_{2} = \frac{h}{\sqrt{2 m e (81)}} = \frac{k}{9}, λ_{3} = \frac{h}{\sqrt{2 m e (49)}} = \frac{k}{7}$

$\frac{λ_{3} - λ_{2}}{λ_{1}} = \frac{\frac{k}{7} - \frac{k}{9}}{\frac{k}{10}} = \frac{(\frac{2 k}{63})}{\frac{k}{10}} = \frac{20}{63}$

Solution

1 $\frac{20}{41}$

2 $\frac{10}{63}$

3 $\frac{20}{63}$

4 $\frac{10}{41}$

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Solution

1 2041

2 1063

3 2063

4 1041

Ans. (3) λ=h2meV, λ1=h2me(100)=h102me=k10, k=h2me λ2=h2me(81)=k9, λ3=h2me(49)=k7 λ3-λ2λ1=k7-k9k10=(2k63)k10=2063

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1 $\frac{20}{41}$

2 $\frac{10}{63}$

3 $\frac{20}{63}$

4 $\frac{10}{41}$