If Planck-s constant is denoted by h and electronic charge by e- then photoelectric effect allows determination of-Only eBoth h and eOnly h-eOnly h

Question

Toppr Admin · Accepted Answer

The correct option is D Only h-eIn photoelectric effect we can know only the stopping potential-Stopping potential -hc-x3BB-x2212-x3D5-e-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -hce-x3BB-x2212-x3D5-eSo- we can determine only h-e because we know stopping potential- frequency and work function of the metal-

If Planck’s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:
Only e
Both h and e
Only h/e
Only h

The correct option is D Only h/e
In photoelectric effect we can know only the stopping potential.
Stopping potential $= (\frac{h c}{λ} - ϕ) / e$
$= \frac{h c}{e λ} - ϕ / e$
So, we can determine only $h / e$ because we know stopping potential, frequency and work function of the metal.

If Planck’s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:Only eBoth h and eOnly h/eOnly h

The correct option is D Only h/eIn photoelectric effect we can know only the stopping potential.Stopping potential =(hcλ−ϕ)/e =hceλ−ϕ/eSo, we can determine only h/e because we know stopping potential, frequency and work function of the metal.

If Planck’s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:
Only e
Both h and e
Only h/e
Only h

The correct option is D Only h/e
In photoelectric effect we can know only the stopping potential.
Stopping potential $= (\frac{h c}{λ} - ϕ) / e$
$= \frac{h c}{e λ} - ϕ / e$
So, we can determine only $h / e$ because we know stopping potential, frequency and work function of the metal.