The work of a metal in 4 eV- For the emission of photoelectrons of zero velocity from the metal surface- the wavelength of the incident radiation should be1700overset -circ-A-2700overset -circ-A-3100overset -circ-A-5900overset -circ-A-

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Toppr Admin · Accepted Answer

The correct option is C 3100-x2218-AWe know-KE-hv-x2212-x3D5-1-2mv2-hc-x3BB-x2212-x3D5-Velocity -012400-x3BB-A0-4eV-x2234-xA0- -xA0-x3BB-124004-3100A0

The work function of a metal in $4 e V$ . For the emission of photoelectrons of zero velocity from the metal surface, the wavelength of the incident radiation should be
$1700 \circ A$
$2700 \circ A$
$3100 \circ A$
$5900 \circ A$

The correct option is C $3100 \circ A$
We know,
$K E = h v - ϕ$
$1 / 2 {m v}^{2} = \frac{h c}{λ} - ϕ$
Velocity $= 0$
$\frac{12400}{λ (A^{0})} = 4 e V$
$∴$ $λ = \frac{12400}{4} = 3100 A^{0}$

The work function of a metal in 4 eV. For the emission of photoelectrons of zero velocity from the metal surface, the wavelength of the incident radiation should be1700∘A2700∘A3100∘A5900∘A

The correct option is C 3100∘AWe know,KE=hv−ϕ1/2mv2=hcλ−ϕVelocity =012400λ(A0)=4eV∴ λ=124004=3100A0

The work function of a metal in $4 e V$ . For the emission of photoelectrons of zero velocity from the metal surface, the wavelength of the incident radiation should be
$1700 \circ A$
$2700 \circ A$
$3100 \circ A$
$5900 \circ A$

The correct option is C $3100 \circ A$
We know,
$K E = h v - ϕ$
$1 / 2 {m v}^{2} = \frac{h c}{λ} - ϕ$
Velocity $= 0$
$\frac{12400}{λ (A^{0})} = 4 e V$
$∴$ $λ = \frac{12400}{4} = 3100 A^{0}$