The de Broglie wavelength associated with an electron accelerated through a potential difference of $\frac{200}{3} \,V$ is nearly (in $Å$)

The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

$A$ proton moves in a circular path of radius $6.6 \times 10^{-3} \ m$ perpendicular to a magnetic field of $0.625 \ T$. Find the de Broglie wavelength associated with the proton in $\mathring{A}$.

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

$A$ wave is associated with matter:

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be: