Answer:
(c) 0.174 nm
Explanation:
According to de Broglie hypothesis, the wavelength of the wave associated with electron is given by:
[tex] \boxed{ \bf{\lambda = \sqrt{\dfrac{150}{V \ (in \ Volt)}} \: \text{\AA}}}[/tex]
V → Potential Difference (50.0 V)
By substituting value of potential difference in the equation, we get:
[tex] \rm \longrightarrow \lambda = \sqrt{\dfrac{150}{50}} \: \text{\AA} \\ \\ \rm \longrightarrow \lambda = \sqrt{3} \: \text{\AA} \\ \\ \rm \longrightarrow \lambda = 1.74 \: \text{\AA} \\ \\ \rm \longrightarrow \lambda = 0.174 \: nm[/tex]