For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) :

Question

For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) : (A) 2 √h v / m (B) √h v / (2m) (C) √h v / m (D)  √2h v / m

Tardigrade Admin · Accepted Answer

h(2 v) = hv + (1/2) mv2max ⇒ = vmax = √(2hv/m)

Q. For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) :

$2 h v / m$

$h v / (2 m)$

$h v / m$

$2 h v / m$

$h (2 v) = h v + \frac{1}{2} m v_{ma x}^{2} \Rightarrow= v_{ma x} = \frac{2 h v}{m}$

Q. For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) :

2hv/m​

hv/(2m)​

hv/m​

2hv/m​

h(2v)=hv+21​mvmax2​⇒=vmax​=m2hv​​

$2 h v / m$

$h v / (2 m)$

$h v / m$

$2 h v / m$

$h (2 v) = h v + \frac{1}{2} m v_{ma x}^{2} \Rightarrow= v_{ma x} = \frac{2 h v}{m}$