The work function of nickel is 5 text{ eV}. W | vedclass

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(C) The stopping potential $V_0$ is given by the Einstein's photoelectric equation: $e V_0 = K_{	ext{max}} = \frac{hc}{\lambda} - \phi$. Given: $h =

Vedclass · Accepted Answer

$1.25$

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(C) The stopping potential $V_0$ is given by the Einstein's photoelectric equation: $e V_0 = K_{	ext{max}} = \frac{hc}{\lambda} - \phi$. Given: $h = 6.67 	imes 10^{-34} 	ext{ J-s}$, $c = 3 	imes 10^8 	ext{ m/s}$, $\lambda = 2000 	ext{ \AA} = 2 	imes 10^{-7} 	ext{ m}$, and $\phi = 5 	ext{ eV} = 5 	imes 1.6 	imes 10^{-19} 	ext{ J} = 8 	imes 10^{-19} 	ext{ J}$. Calculating the energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{6.67 	imes 10^{-34} 	imes 3 	imes 10^8}{2 	imes 10^{-7}} = 10.005 	imes 10^{-19} 	ext{ J}$. Substituting into the equation: $e V_0 = 10.005 	imes 10^{-19} 	ext{ J} - 8 	imes 10^{-19} 	ext{ J} = 2.005 	imes 10^{-19} 	ext{ J}$. Therefore, $V_0 = \frac{2.005 	imes 10^{-19} 	ext{ J}}{1.6 	imes 10^{-19} 	ext{ C}} \approx 1.25 	ext{ V}$.

List$-I$	List$-II$
$(P)$ image	$(1)$ Reaction with phenyl diazonium salt gives yellow dye.
$(Q)$ image	$(2)$ Reaction with ninhydrin gives purple color and it also reacts with $FeCl _3$ to give violet color.
$(R)$ image	$(3)$ Reaction with glucose will give corresponding hydrazone.
$(S)$ image	$(4)$ Lassiagne extract of the compound treated with dilute HCl followed by addition of aqueous $FeCl _3$ gives blood red color.
	$(5)$ After complete hydrolysis, it will give ninhydrin test and it $\text{DOES NOT}$ give positive phthalein dye test.

The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ \AA}$ falls on it, it emits photoelectrons. The potential difference necessary to stop the fastest electrons emitted is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

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