The reason for current flow in a $P-N$ junction in forward bias is

In the circuit shown, assume the diode to be ideal. When $V_{i}$ increases from $2 \ V$ to $6 \ V$, the change in the current is (in $mA$):

The ratio of the resistance of a $P-N$ junction diode in forward bias to its resistance in reverse bias is approximately:

The energy of a photon in a monochromatic light of wavelength $621 \ nm$ matches with the band gap of a semiconducting material. Then the minimum energy required to create an electron-hole pair from the semiconductor is
[Take $hc = 1242 \ eV-nm$,where $h$ is Planck's constant and $c$ is speed of light in vacuum] (in $eV$)

In the following circuit,find $I_1$ and $I_2$.

When a $P-N$ junction is reverse biased,then ...